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Popular Trigonometria >

cosh(4x)=16sinh(x)+1

Solução

cosh (4 x) = 16 sinh (x) + 1

Solução

x = 0, x = ln (2.41421 \dots)

Graus

x = 0^{\circ}, x = 50.49898 \dots^{\circ}

Passos da solução

cosh (4 x) = 16 sinh (x) + 1

Reeecreva usando identidades trigonométricas

cosh (4 x) = 16 sinh (x) + 1

Use a identidade hiperbólica:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

cosh (4 x) = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} + 1

Use a identidade hiperbólica:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} + 1

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} + 1

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} + 1 : x = 0, x = ln (2.41421 \dots)

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} + 1

Multiplicar ambos os lados por

2

\frac{e ^{4 x} + e ^{- 4 x}}{2} \cdot 2 = 16 \cdot \frac{e ^{x} - e ^{- x}}{2} \cdot 2 + 1 \cdot 2

Simplificar

e^{4 x} + e^{- 4 x} = 16 (e^{x} - e^{- x}) + 2

Aplicar as propriedades dos expoentes

e^{4 x} + e^{- 4 x} = 16 (e^{x} - e^{- x}) + 2

Aplicar as propriedades dos expoentes:

a^{b c} = (a^{b})^{c}

e^{4 x} = (e^{x})^{4}, e^{- 4 x} = (e^{x})^{- 4}, e^{- x} = (e^{x})^{- 1}

(e^{x})^{4} + (e^{x})^{- 4} = 16 (e^{x} - (e^{x})^{- 1}) + 2

(e^{x})^{4} + (e^{x})^{- 4} = 16 (e^{x} - (e^{x})^{- 1}) + 2

Reescrever a equação com

e^{x} = u

(u)^{4} + (u)^{- 4} = 16 (u - (u)^{- 1}) + 2

Resolver

u^{4} + u^{- 4} = 16 (u - u^{- 1}) + 2 : u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

u^{4} + u^{- 4} = 16 (u - u^{- 1}) + 2

Simplificar

u^{4} + \frac{1}{u ^{4}} = 16 (u - \frac{1}{u}) + 2

Multiplicar ambos os lados por

u^{4}

u^{4} + \frac{1}{u ^{4}} = 16 (u - \frac{1}{u}) + 2

Multiplicar ambos os lados por

u^{4}

u^{4} u^{4} + \frac{1}{u ^{4}} u^{4} = 16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

Simplificar

u^{4} u^{4} + \frac{1}{u ^{4}} u^{4} = 16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

Simplificar

u^{4} u^{4} : u^{8}

u^{4} u^{4}

Aplicar as propriedades dos expoentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{4} u^{4} = u^{4 + 4}

= u^{4 + 4}

Somar:

4 + 4 = 8

= u^{8}

Simplificar

\frac{1}{u ^{4}} u^{4} : 1

\frac{1}{u ^{4}} u^{4}

Multiplicar frações:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot u ^{4}}{u ^{4}}

Eliminar o fator comum:

u^{4}

= 1

u^{8} + 1 = 16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

u^{8} + 1 = 16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

u^{8} + 1 = 16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

Expandir

16 (u - \frac{1}{u}) u^{4} + 2 u^{4} : 16 u^{5} - 16 u^{3} + 2 u^{4}

16 (u - \frac{1}{u}) u^{4} + 2 u^{4}

= 16 u^{4} (u - \frac{1}{u}) + 2 u^{4}

Expandir

16 u^{4} (u - \frac{1}{u}) : 16 u^{5} - 16 u^{3}

16 u^{4} (u - \frac{1}{u})

Colocar os parênteses utilizando:

a (b - c) = ab - a c

a = 16 u^{4}, b = u, c = \frac{1}{u}

= 16 u^{4} u - 16 u^{4} \frac{1}{u}

= 16 u^{4} u - 16 \cdot \frac{1}{u} u^{4}

Simplificar

16 u^{4} u - 16 \cdot \frac{1}{u} u^{4} : 16 u^{5} - 16 u^{3}

16 u^{4} u - 16 \cdot \frac{1}{u} u^{4}

16 u^{4} u = 16 u^{5}

16 u^{4} u

Aplicar as propriedades dos expoentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{4} u = u^{4 + 1}

= 16 u^{4 + 1}

Somar:

4 + 1 = 5

= 16 u^{5}

16 \cdot \frac{1}{u} u^{4} = 16 u^{3}

16 \cdot \frac{1}{u} u^{4}

Multiplicar frações:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 16 u ^{4}}{u}

Multiplicar os números:

1 \cdot 16 = 16

= \frac{16 u ^{4}}{u}

Eliminar o fator comum:

u

= 16 u^{3}

= 16 u^{5} - 16 u^{3}

= 16 u^{5} - 16 u^{3}

= 16 u^{5} - 16 u^{3} + 2 u^{4}

u^{8} + 1 = 16 u^{5} - 16 u^{3} + 2 u^{4}

Resolver

u^{8} + 1 = 16 u^{5} - 16 u^{3} + 2 u^{4} : u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

u^{8} + 1 = 16 u^{5} - 16 u^{3} + 2 u^{4}

Mova

2 u^{4}

para o lado esquerdo

u^{8} + 1 = 16 u^{5} - 16 u^{3} + 2 u^{4}

Subtrair

2 u^{4}

de ambos os lados

u^{8} + 1 - 2 u^{4} = 16 u^{5} - 16 u^{3} + 2 u^{4} - 2 u^{4}

Simplificar

u^{8} + 1 - 2 u^{4} = 16 u^{5} - 16 u^{3}

u^{8} + 1 - 2 u^{4} = 16 u^{5} - 16 u^{3}

Mova

16 u^{3}

para o lado esquerdo

u^{8} + 1 - 2 u^{4} = 16 u^{5} - 16 u^{3}

Adicionar

16 u^{3}

a ambos os lados

u^{8} + 1 - 2 u^{4} + 16 u^{3} = 16 u^{5} - 16 u^{3} + 16 u^{3}

Simplificar

u^{8} + 1 - 2 u^{4} + 16 u^{3} = 16 u^{5}

u^{8} + 1 - 2 u^{4} + 16 u^{3} = 16 u^{5}

Mova

16 u^{5}

para o lado esquerdo

u^{8} + 1 - 2 u^{4} + 16 u^{3} = 16 u^{5}

Subtrair

16 u^{5}

de ambos os lados

u^{8} + 1 - 2 u^{4} + 16 u^{3} - 16 u^{5} = 16 u^{5} - 16 u^{5}

Simplificar

u^{8} + 1 - 2 u^{4} + 16 u^{3} - 16 u^{5} = 0

u^{8} + 1 - 2 u^{4} + 16 u^{3} - 16 u^{5} = 0

Escrever na forma padrão

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{8} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1 = 0

Fatorar

u^{8} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1 : (u + 1) (u - 1) (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1)

u^{8} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

Utilizar o teorema das raízes racionais

a_{0} = 1, a_{n} = 1

Os divisores de

a_{0} : 1,

Os divisores de

a_{n} : 1

Portanto, verificar os seguintes números racionais:

\pm \frac{1}{1}

- \frac{1}{1}

é a raiz da expressão, portanto, fatorar

u + 1

= (u + 1) \frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

\frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

\frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir

\frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} : \frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = u^{7} + \frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

u^{8} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

e o divisor

u + 1 : \frac{u ^{8}}{u} = u^{7}

Quociente = u^{7}

Multiplicar

u + 1

por

u^{7} : u^{8} + u^{7}

Subtrair

u^{8} + u^{7}

u^{8} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

para obter um novo resto

Resto = - u^{7} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

Portanto

\frac{u ^{8} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = u^{7} + \frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

= u^{7} + \frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir

\frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} : \frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = - u^{6} + \frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

- u^{7} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

e o divisor

u + 1 : \frac{- u ^{7}}{u} = - u^{6}

Quociente = - u^{6}

Multiplicar

u + 1

por

- u^{6} : - u^{7} - u^{6}

Subtrair

- u^{7} - u^{6}

- u^{7} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

para obter um novo resto

Resto = u^{6} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

Portanto

\frac{- u ^{7} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = - u^{6} + \frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

= u^{7} - u^{6} + \frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir

\frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} : \frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = u^{5} + \frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

u^{6} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

e o divisor

u + 1 : \frac{u ^{6}}{u} = u^{5}

Quociente = u^{5}

Multiplicar

u + 1

por

u^{5} : u^{6} + u^{5}

Subtrair

u^{6} + u^{5}

u^{6} - 16 u^{5} - 2 u^{4} + 16 u^{3} + 1

para obter um novo resto

Resto = - 17 u^{5} - 2 u^{4} + 16 u^{3} + 1

Portanto

\frac{u ^{6} - 16 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = u^{5} + \frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

= u^{7} - u^{6} + u^{5} + \frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir

\frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} : \frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = - 17 u^{4} + \frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

- 17 u^{5} - 2 u^{4} + 16 u^{3} + 1

e o divisor

u + 1 : \frac{- 17 u ^{5}}{u} = - 17 u^{4}

Quociente = - 17 u^{4}

Multiplicar

u + 1

por

- 17 u^{4} : - 17 u^{5} - 17 u^{4}

Subtrair

- 17 u^{5} - 17 u^{4}

- 17 u^{5} - 2 u^{4} + 16 u^{3} + 1

para obter um novo resto

Resto = 15 u^{4} + 16 u^{3} + 1

Portanto

\frac{- 17 u ^{5} - 2 u ^{4} + 16 u ^{3} + 1}{u + 1} = - 17 u^{4} + \frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1}

= u^{7} - u^{6} + u^{5} - 17 u^{4} + \frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1}

Dividir

\frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1} : \frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1} = 15 u^{3} + \frac{u ^{3} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

15 u^{4} + 16 u^{3} + 1

e o divisor

u + 1 : \frac{15 u ^{4}}{u} = 15 u^{3}

Quociente = 15 u^{3}

Multiplicar

u + 1

por

15 u^{3} : 15 u^{4} + 15 u^{3}

Subtrair

15 u^{4} + 15 u^{3}

15 u^{4} + 16 u^{3} + 1

para obter um novo resto

Resto = u^{3} + 1

Portanto

\frac{15 u ^{4} + 16 u ^{3} + 1}{u + 1} = 15 u^{3} + \frac{u ^{3} + 1}{u + 1}

= u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + \frac{u ^{3} + 1}{u + 1}

Dividir

\frac{u ^{3} + 1}{u + 1} : \frac{u ^{3} + 1}{u + 1} = u^{2} + \frac{- u ^{2} + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

u^{3} + 1

e o divisor

u + 1 : \frac{u ^{3}}{u} = u^{2}

Quociente = u^{2}

Multiplicar

u + 1

por

u^{2} : u^{3} + u^{2}

Subtrair

u^{3} + u^{2}

u^{3} + 1

para obter um novo resto

Resto = - u^{2} + 1

Portanto

\frac{u ^{3} + 1}{u + 1} = u^{2} + \frac{- u ^{2} + 1}{u + 1}

= u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} + \frac{- u ^{2} + 1}{u + 1}

Dividir

\frac{- u ^{2} + 1}{u + 1} : \frac{- u ^{2} + 1}{u + 1} = - u + \frac{u + 1}{u + 1}

Dividir os coeficientes dos termos de maior grau do numerador

- u^{2} + 1

e o divisor

u + 1 : \frac{- u ^{2}}{u} = - u

Quociente = - u

Multiplicar

u + 1

por

- u : - u^{2} - u

Subtrair

- u^{2} - u

- u^{2} + 1

para obter um novo resto

Resto = u + 1

Portanto

\frac{- u ^{2} + 1}{u + 1} = - u + \frac{u + 1}{u + 1}

= u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + \frac{u + 1}{u + 1}

Dividir

\frac{u + 1}{u + 1} : \frac{u + 1}{u + 1} = 1

Dividir os coeficientes dos termos de maior grau do numerador

u + 1

e o divisor

u + 1 : \frac{u}{u} = 1

Quociente = 1

Multiplicar

u + 1

por

1 : u + 1

Subtrair

u + 1

u + 1

para obter um novo resto

Resto = 0

Portanto

\frac{u + 1}{u + 1} = 1

= u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

= u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

Fatorar

u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1 : (u - 1) (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1)

u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

Utilizar o teorema das raízes racionais

a_{0} = 1, a_{n} = 1

Os divisores de

a_{0} : 1,

Os divisores de

a_{n} : 1

Portanto, verificar os seguintes números racionais:

\pm \frac{1}{1}

\frac{1}{1}

é a raiz da expressão, portanto, fatorar

u - 1

= (u - 1) \frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

\frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = u^{6} + u^{4} - 16 u^{3} - u^{2} - 1

\frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir

\frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} : \frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = u^{6} + \frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir os coeficientes dos termos de maior grau do numerador

u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

e o divisor

u - 1 : \frac{u ^{7}}{u} = u^{6}

Quociente = u^{6}

Multiplicar

u - 1

por

u^{6} : u^{7} - u^{6}

Subtrair

u^{7} - u^{6}

u^{7} - u^{6} + u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

para obter um novo resto

Resto = u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

Portanto

\frac{u ^{7} - u ^{6} + u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = u^{6} + \frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

= u^{6} + \frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir

\frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} : \frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = u^{4} + \frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir os coeficientes dos termos de maior grau do numerador

u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

e o divisor

u - 1 : \frac{u ^{5}}{u} = u^{4}

Quociente = u^{4}

Multiplicar

u - 1

por

u^{4} : u^{5} - u^{4}

Subtrair

u^{5} - u^{4}

u^{5} - 17 u^{4} + 15 u^{3} + u^{2} - u + 1

para obter um novo resto

Resto = - 16 u^{4} + 15 u^{3} + u^{2} - u + 1

Portanto

\frac{u ^{5} - 17 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = u^{4} + \frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

= u^{6} + u^{4} + \frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir

\frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} : \frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = - 16 u^{3} + \frac{- u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir os coeficientes dos termos de maior grau do numerador

- 16 u^{4} + 15 u^{3} + u^{2} - u + 1

e o divisor

u - 1 : \frac{- 16 u ^{4}}{u} = - 16 u^{3}

Quociente = - 16 u^{3}

Multiplicar

u - 1

por

- 16 u^{3} : - 16 u^{4} + 16 u^{3}

Subtrair

- 16 u^{4} + 16 u^{3}

- 16 u^{4} + 15 u^{3} + u^{2} - u + 1

para obter um novo resto

Resto = - u^{3} + u^{2} - u + 1

Portanto

\frac{- 16 u ^{4} + 15 u ^{3} + u ^{2} - u + 1}{u - 1} = - 16 u^{3} + \frac{- u ^{3} + u ^{2} - u + 1}{u - 1}

= u^{6} + u^{4} - 16 u^{3} + \frac{- u ^{3} + u ^{2} - u + 1}{u - 1}

Dividir

\frac{- u ^{3} + u ^{2} - u + 1}{u - 1} : \frac{- u ^{3} + u ^{2} - u + 1}{u - 1} = - u^{2} + \frac{- u + 1}{u - 1}

Dividir os coeficientes dos termos de maior grau do numerador

- u^{3} + u^{2} - u + 1

e o divisor

u - 1 : \frac{- u ^{3}}{u} = - u^{2}

Quociente = - u^{2}

Multiplicar

u - 1

por

- u^{2} : - u^{3} + u^{2}

Subtrair

- u^{3} + u^{2}

- u^{3} + u^{2} - u + 1

para obter um novo resto

Resto = - u + 1

Portanto

\frac{- u ^{3} + u ^{2} - u + 1}{u - 1} = - u^{2} + \frac{- u + 1}{u - 1}

= u^{6} + u^{4} - 16 u^{3} - u^{2} + \frac{- u + 1}{u - 1}

Dividir

\frac{- u + 1}{u - 1} : \frac{- u + 1}{u - 1} = - 1

Dividir os coeficientes dos termos de maior grau do numerador

- u + 1

e o divisor

u - 1 : \frac{- u}{u} = - 1

Quociente = - 1

Multiplicar

u - 1

por

- 1 : - u + 1

Subtrair

- u + 1

- u + 1

para obter um novo resto

Resto = 0

Portanto

\frac{- u + 1}{u - 1} = - 1

= u^{6} + u^{4} - 16 u^{3} - u^{2} - 1

= u^{6} + u^{4} - 16 u^{3} - u^{2} - 1

= (u - 1) (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1)

= (u + 1) (u - 1) (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1)

(u + 1) (u - 1) (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1) = 0

Usando o princípio do fator zero: Se

ab = 0

então

a = 0

b = 0

u + 1 = 0 or u - 1 = 0 or u^{6} + u^{4} - 16 u^{3} - u^{2} - 1 = 0

Resolver

u + 1 = 0 : u = - 1

u + 1 = 0

Mova

1

para o lado direito

u + 1 = 0

Subtrair

1

de ambos os lados

u + 1 - 1 = 0 - 1

Simplificar

u = - 1

u = - 1

Resolver

u - 1 = 0 : u = 1

u - 1 = 0

Mova

1

para o lado direito

u - 1 = 0

Adicionar

1

a ambos os lados

u - 1 + 1 = 0 + 1

Simplificar

u = 1

u = 1

Resolver

u^{6} + u^{4} - 16 u^{3} - u^{2} - 1 = 0 : u \approx - 0.41421 \dots, u \approx 2.41421 \dots

u^{6} + u^{4} - 16 u^{3} - u^{2} - 1 = 0

Encontrar uma solução para

u^{6} + u^{4} - 16 u^{3} - u^{2} - 1 = 0

utilizando o método de Newton-Raphson:

u \approx - 0.41421 \dots

u^{6} + u^{4} - 16 u^{3} - u^{2} - 1 = 0

Definição de método de Newton-Raphson

f (u) = u^{6} + u^{4} - 16 u^{3} - u^{2} - 1

Encontrar

f^{^{'}} (u) : 6 u^{5} + 4 u^{3} - 48 u^{2} - 2 u

\frac{d}{d u} (u^{6} + u^{4} - 16 u^{3} - u^{2} - 1)

Aplicar a regra da soma/diferença:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{6}) + \frac{d}{d u} (u^{4}) - \frac{d}{d u} (16 u^{3}) - \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{6}) = 6 u^{5}

\frac{d}{d u} (u^{6})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 u^{6 - 1}

Simplificar

= 6 u^{5}

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 u^{4 - 1}

Simplificar

= 4 u^{3}

\frac{d}{d u} (16 u^{3}) = 48 u^{2}

\frac{d}{d u} (16 u^{3})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 16 \frac{d}{d u} (u^{3})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 16 \cdot 3 u^{3 - 1}

Simplificar

= 48 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplificar

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivada de uma constante:

\frac{d}{d x} (a) = 0

= 0

= 6 u^{5} + 4 u^{3} - 48 u^{2} - 2 u - 0

Simplificar

= 6 u^{5} + 4 u^{3} - 48 u^{2} - 2 u

Seja

u_{0} = 1

Calcular

u_{n + 1}

até que

Δ u_{n + 1} < 0.000001

u_{1} = 0.6 : Δ u_{1} = 0.4

f (u_{0}) = 1^{6} + 1^{4} - 16 \cdot 1^{3} - 1^{2} - 1 = - 16

f^{^{'}} (u_{0}) = 6 \cdot 1^{5} + 4 \cdot 1^{3} - 48 \cdot 1^{2} - 2 \cdot 1 = - 40

u_{1} = 0.6

Δ u_{1} = ∣ 0.6 - 1 ∣ = 0.4

Δ u_{1} = 0.4

u_{2} = 0.32945 \dots : Δ u_{2} = 0.27054 \dots

f (u_{1}) = 0. 6^{6} + 0. 6^{4} - 16 \cdot 0. 6^{3} - 0. 6^{2} - 1 = - 4.639744

f^{^{'}} (u_{1}) = 6 \cdot 0. 6^{5} + 4 \cdot 0. 6^{3} - 48 \cdot 0. 6^{2} - 2 \cdot 0.6 = - 17.14944

u_{2} = 0.32945 \dots

Δ u_{2} = ∣ 0.32945 \dots - 0.6 ∣ = 0.27054 \dots

Δ u_{2} = 0.27054 \dots

u_{3} = 0.03701 \dots : Δ u_{3} = 0.29243 \dots

f (u_{2}) = 0.32945 \dots^{6} + 0.32945 \dots^{4} - 16 \cdot 0.32945 \dots^{3} - 0.32945 \dots^{2} - 1 = - 1.66761 \dots

f^{^{'}} (u_{2}) = 6 \cdot 0.32945 \dots^{5} + 4 \cdot 0.32945 \dots^{3} - 48 \cdot 0.32945 \dots^{2} - 2 \cdot 0.32945 \dots = - 5.70244 \dots

u_{3} = 0.03701 \dots

Δ u_{3} = ∣ 0.03701 \dots - 0.32945 \dots ∣ = 0.29243 \dots

Δ u_{3} = 0.29243 \dots

u_{4} = - 7.14264 \dots : Δ u_{4} = 7.17966 \dots

f (u_{3}) = 0.03701 \dots^{6} + 0.03701 \dots^{4} - 16 \cdot 0.03701 \dots^{3} - 0.03701 \dots^{2} - 1 = - 1.00217 \dots

f^{^{'}} (u_{3}) = 6 \cdot 0.03701 \dots^{5} + 4 \cdot 0.03701 \dots^{3} - 48 \cdot 0.03701 \dots^{2} - 2 \cdot 0.03701 \dots = - 0.13958 \dots

u_{4} = - 7.14264 \dots

Δ u_{4} = ∣ - 7.14264 \dots - 0.03701 \dots ∣ = 7.17966 \dots

Δ u_{4} = 7.17966 \dots

u_{5} = - 5.91974 \dots : Δ u_{5} = 1.22290 \dots

f (u_{4}) = (- 7.14264 \dots)^{6} + (- 7.14264 \dots)^{4} - 16 (- 7.14264 \dots)^{3} - (- 7.14264 \dots)^{2} - 1 = 141168.16963 \dots

f^{^{'}} (u_{4}) = 6 (- 7.14264 \dots)^{5} + 4 (- 7.14264 \dots)^{3} - 48 (- 7.14264 \dots)^{2} - 2 (- 7.14264 \dots) = - 115436.50403 \dots

u_{5} = - 5.91974 \dots

Δ u_{5} = ∣ - 5.91974 \dots - (- 7.14264 \dots) ∣ = 1.22290 \dots

Δ u_{5} = 1.22290 \dots

u_{6} = - 4.88878 \dots : Δ u_{6} = 1.03095 \dots

f (u_{5}) = (- 5.91974 \dots)^{6} + (- 5.91974 \dots)^{4} - 16 (- 5.91974 \dots)^{3} - (- 5.91974 \dots)^{2} - 1 = 47545.59081 \dots

f^{^{'}} (u_{5}) = 6 (- 5.91974 \dots)^{5} + 4 (- 5.91974 \dots)^{3} - 48 (- 5.91974 \dots)^{2} - 2 (- 5.91974 \dots) = - 46117.92631 \dots

u_{6} = - 4.88878 \dots

Δ u_{6} = ∣ - 4.88878 \dots - (- 5.91974 \dots) ∣ = 1.03095 \dots

Δ u_{6} = 1.03095 \dots

u_{7} = - 4.01362 \dots : Δ u_{7} = 0.87515 \dots

f (u_{6}) = (- 4.88878 \dots)^{6} + (- 4.88878 \dots)^{4} - 16 (- 4.88878 \dots)^{3} - (- 4.88878 \dots)^{2} - 1 = 16068.08422 \dots

f^{^{'}} (u_{6}) = 6 (- 4.88878 \dots)^{5} + 4 (- 4.88878 \dots)^{3} - 48 (- 4.88878 \dots)^{2} - 2 (- 4.88878 \dots) = - 18360.23113 \dots

u_{7} = - 4.01362 \dots

Δ u_{7} = ∣ - 4.01362 \dots - (- 4.88878 \dots) ∣ = 0.87515 \dots

Δ u_{7} = 0.87515 \dots

u_{8} = - 3.26329 \dots : Δ u_{8} = 0.75033 \dots

f (u_{7}) = (- 4.01362 \dots)^{6} + (- 4.01362 \dots)^{4} - 16 (- 4.01362 \dots)^{3} - (- 4.01362 \dots)^{2} - 1 = 5457.33921 \dots

f^{^{'}} (u_{7}) = 6 (- 4.01362 \dots)^{5} + 4 (- 4.01362 \dots)^{3} - 48 (- 4.01362 \dots)^{2} - 2 (- 4.01362 \dots) = - 7273.21083 \dots

u_{8} = - 3.26329 \dots

Δ u_{8} = ∣ - 3.26329 \dots - (- 4.01362 \dots) ∣ = 0.75033 \dots

Δ u_{8} = 0.75033 \dots

u_{9} = - 2.61197 \dots : Δ u_{9} = 0.65132 \dots

f (u_{8}) = (- 3.26329 \dots)^{6} + (- 3.26329 \dots)^{4} - 16 (- 3.26329 \dots)^{3} - (- 3.26329 \dots)^{2} - 1 = 1865.40737 \dots

f^{^{'}} (u_{8}) = 6 (- 3.26329 \dots)^{5} + 4 (- 3.26329 \dots)^{3} - 48 (- 3.26329 \dots)^{2} - 2 (- 3.26329 \dots) = - 2864.03467 \dots

u_{9} = - 2.61197 \dots

Δ u_{9} = ∣ - 2.61197 \dots - (- 3.26329 \dots) ∣ = 0.65132 \dots

Δ u_{9} = 0.65132 \dots

u_{10} = - 2.04081 \dots : Δ u_{10} = 0.57115 \dots

f (u_{9}) = (- 2.61197 \dots)^{6} + (- 2.61197 \dots)^{4} - 16 (- 2.61197 \dots)^{3} - (- 2.61197 \dots)^{2} - 1 = 641.38974 \dots

f^{^{'}} (u_{9}) = 6 (- 2.61197 \dots)^{5} + 4 (- 2.61197 \dots)^{3} - 48 (- 2.61197 \dots)^{2} - 2 (- 2.61197 \dots) = - 1122.97668 \dots

u_{10} = - 2.04081 \dots

Δ u_{10} = ∣ - 2.04081 \dots - (- 2.61197 \dots) ∣ = 0.57115 \dots

Δ u_{10} = 0.57115 \dots

u_{11} = - 1.54238 \dots : Δ u_{11} = 0.49843 \dots

f (u_{10}) = (- 2.04081 \dots)^{6} + (- 2.04081 \dots)^{4} - 16 (- 2.04081 \dots)^{3} - (- 2.04081 \dots)^{2} - 1 = 220.42864 \dots

f^{^{'}} (u_{10}) = 6 (- 2.04081 \dots)^{5} + 4 (- 2.04081 \dots)^{3} - 48 (- 2.04081 \dots)^{2} - 2 (- 2.04081 \dots) = - 442.24518 \dots

u_{11} = - 1.54238 \dots

Δ u_{11} = ∣ - 1.54238 \dots - (- 2.04081 \dots) ∣ = 0.49843 \dots

Δ u_{11} = 0.49843 \dots

u_{12} = - 1.12448 \dots : Δ u_{12} = 0.41790 \dots

f (u_{11}) = (- 1.54238 \dots)^{6} + (- 1.54238 \dots)^{4} - 16 (- 1.54238 \dots)^{3} - (- 1.54238 \dots)^{2} - 1 = 74.45277 \dots

f^{^{'}} (u_{11}) = 6 (- 1.54238 \dots)^{5} + 4 (- 1.54238 \dots)^{3} - 48 (- 1.54238 \dots)^{2} - 2 (- 1.54238 \dots) = - 178.15724 \dots

u_{12} = - 1.12448 \dots

Δ u_{12} = ∣ - 1.12448 \dots - (- 1.54238 \dots) ∣ = 0.41790 \dots

Δ u_{12} = 0.41790 \dots

u_{13} = - 0.80272 \dots : Δ u_{13} = 0.32175 \dots

f (u_{12}) = (- 1.12448 \dots)^{6} + (- 1.12448 \dots)^{4} - 16 (- 1.12448 \dots)^{3} - (- 1.12448 \dots)^{2} - 1 = 24.10604 \dots

f^{^{'}} (u_{12}) = 6 (- 1.12448 \dots)^{5} + 4 (- 1.12448 \dots)^{3} - 48 (- 1.12448 \dots)^{2} - 2 (- 1.12448 \dots) = - 74.92024 \dots

u_{13} = - 0.80272 \dots

Δ u_{13} = ∣ - 0.80272 \dots - (- 1.12448 \dots) ∣ = 0.32175 \dots

Δ u_{13} = 0.32175 \dots

u_{14} = - 0.58368 \dots : Δ u_{14} = 0.21904 \dots

f (u_{13}) = (- 0.80272 \dots)^{6} + (- 0.80272 \dots)^{4} - 16 (- 0.80272 \dots)^{3} - (- 0.80272 \dots)^{2} - 1 = 7.31448 \dots

f^{^{'}} (u_{13}) = 6 (- 0.80272 \dots)^{5} + 4 (- 0.80272 \dots)^{3} - 48 (- 0.80272 \dots)^{2} - 2 (- 0.80272 \dots) = - 33.39326 \dots

u_{14} = - 0.58368 \dots

Δ u_{14} = ∣ - 0.58368 \dots - (- 0.80272 \dots) ∣ = 0.21904 \dots

Δ u_{14} = 0.21904 \dots

u_{15} = - 0.46184 \dots : Δ u_{15} = 0.12183 \dots

f (u_{14}) = (- 0.58368 \dots)^{6} + (- 0.58368 \dots)^{4} - 16 (- 0.58368 \dots)^{3} - (- 0.58368 \dots)^{2} - 1 = 1.99663 \dots

f^{^{'}} (u_{14}) = 6 (- 0.58368 \dots)^{5} + 4 (- 0.58368 \dots)^{3} - 48 (- 0.58368 \dots)^{2} - 2 (- 0.58368 \dots) = - 16.38770 \dots

u_{15} = - 0.46184 \dots

Δ u_{15} = ∣ - 0.46184 \dots - (- 0.58368 \dots) ∣ = 0.12183 \dots

Δ u_{15} = 0.12183 \dots

u_{16} = - 0.41933 \dots : Δ u_{16} = 0.04251 \dots

f (u_{15}) = (- 0.46184 \dots)^{6} + (- 0.46184 \dots)^{4} - 16 (- 0.46184 \dots)^{3} - (- 0.46184 \dots)^{2} - 1 = 0.41814 \dots

f^{^{'}} (u_{15}) = 6 (- 0.46184 \dots)^{5} + 4 (- 0.46184 \dots)^{3} - 48 (- 0.46184 \dots)^{2} - 2 (- 0.46184 \dots) = - 9.83510 \dots

u_{16} = - 0.41933 \dots

Δ u_{16} = ∣ - 0.41933 \dots - (- 0.46184 \dots) ∣ = 0.04251 \dots

Δ u_{16} = 0.04251 \dots

u_{17} = - 0.41428 \dots : Δ u_{17} = 0.00505 \dots

f (u_{16}) = (- 0.41933 \dots)^{6} + (- 0.41933 \dots)^{4} - 16 (- 0.41933 \dots)^{3} - (- 0.41933 \dots)^{2} - 1 = 0.04030 \dots

f^{^{'}} (u_{16}) = 6 (- 0.41933 \dots)^{5} + 4 (- 0.41933 \dots)^{3} - 48 (- 0.41933 \dots)^{2} - 2 (- 0.41933 \dots) = - 7.97447 \dots

u_{17} = - 0.41428 \dots

Δ u_{17} = ∣ - 0.41428 \dots - (- 0.41933 \dots) ∣ = 0.00505 \dots

Δ u_{17} = 0.00505 \dots

u_{18} = - 0.41421 \dots : Δ u_{18} = 0.00006 \dots

f (u_{17}) = (- 0.41428 \dots)^{6} + (- 0.41428 \dots)^{4} - 16 (- 0.41428 \dots)^{3} - (- 0.41428 \dots)^{2} - 1 = 0.00052 \dots

f^{^{'}} (u_{17}) = 6 (- 0.41428 \dots)^{5} + 4 (- 0.41428 \dots)^{3} - 48 (- 0.41428 \dots)^{2} - 2 (- 0.41428 \dots) = - 7.76725 \dots

u_{18} = - 0.41421 \dots

Δ u_{18} = ∣ - 0.41421 \dots - (- 0.41428 \dots) ∣ = 0.00006 \dots

Δ u_{18} = 0.00006 \dots

u_{19} = - 0.41421 \dots : Δ u_{19} = 1.19705 E - 8

f (u_{18}) = (- 0.41421 \dots)^{6} + (- 0.41421 \dots)^{4} - 16 (- 0.41421 \dots)^{3} - (- 0.41421 \dots)^{2} - 1 = 9.29452 E - 8

f^{^{'}} (u_{18}) = 6 (- 0.41421 \dots)^{5} + 4 (- 0.41421 \dots)^{3} - 48 (- 0.41421 \dots)^{2} - 2 (- 0.41421 \dots) = - 7.76450 \dots

u_{19} = - 0.41421 \dots

Δ u_{19} = ∣ - 0.41421 \dots - (- 0.41421 \dots) ∣ = 1.19705 E - 8

Δ u_{19} = 1.19705 E - 8

u \approx - 0.41421 \dots

Aplicar a divisão longa

Eq u a t i o n 0 : \frac{u ^{6} + u ^{4} - 16 u ^{3} - u ^{2} - 1}{u + 0.41421 \dots} = u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots

u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots \approx 0

Encontrar uma solução para

u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots = 0

utilizando o método de Newton-Raphson:

u \approx 2.41421 \dots

u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots = 0

Definição de método de Newton-Raphson

f (u) = u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots

Encontrar

f^{^{'}} (u) : 5 u^{4} - 1.65685 \dots u^{3} + 3.51471 \dots u^{2} - 32.97056 \dots u + 5.82842 \dots

\frac{d}{d u} (u^{5} - 0.41421 \dots u^{4} + 1.17157 \dots u^{3} - 16.48528 \dots u^{2} + 5.82842 \dots u - 2.41421 \dots)

Aplicar a regra da soma/diferença:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{5}) - \frac{d}{d u} (0.41421 \dots u^{4}) + \frac{d}{d u} (1.17157 \dots u^{3}) - \frac{d}{d u} (16.48528 \dots u^{2}) + \frac{d}{d u} (5.82842 \dots u) - \frac{d}{d u} (2.41421 \dots)

\frac{d}{d u} (u^{5}) = 5 u^{4}

\frac{d}{d u} (u^{5})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 u^{5 - 1}

Simplificar

= 5 u^{4}

\frac{d}{d u} (0.41421 \dots u^{4}) = 1.65685 \dots u^{3}

\frac{d}{d u} (0.41421 \dots u^{4})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.41421 \dots \frac{d}{d u} (u^{4})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 0.41421 \dots \cdot 4 u^{4 - 1}

Simplificar

= 1.65685 \dots u^{3}

\frac{d}{d u} (1.17157 \dots u^{3}) = 3.51471 \dots u^{2}

\frac{d}{d u} (1.17157 \dots u^{3})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 1.17157 \dots \frac{d}{d u} (u^{3})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 1.17157 \dots \cdot 3 u^{3 - 1}

Simplificar

= 3.51471 \dots u^{2}

\frac{d}{d u} (16.48528 \dots u^{2}) = 32.97056 \dots u

\frac{d}{d u} (16.48528 \dots u^{2})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 16.48528 \dots \frac{d}{d u} (u^{2})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 16.48528 \dots \cdot 2 u^{2 - 1}

Simplificar

= 32.97056 \dots u

\frac{d}{d u} (5.82842 \dots u) = 5.82842 \dots

\frac{d}{d u} (5.82842 \dots u)

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5.82842 \dots \frac{d u}{d u}

Aplicar a regra da derivação:

\frac{d u}{d u} = 1

= 5.82842 \dots \cdot 1

Simplificar

= 5.82842 \dots

\frac{d}{d u} (2.41421 \dots) = 0

\frac{d}{d u} (2.41421 \dots)

Derivada de uma constante:

\frac{d}{d x} (a) = 0

= 0

= 5 u^{4} - 1.65685 \dots u^{3} + 3.51471 \dots u^{2} - 32.97056 \dots u + 5.82842 \dots - 0

Simplificar

= 5 u^{4} - 1.65685 \dots u^{3} + 3.51471 \dots u^{2} - 32.97056 \dots u + 5.82842 \dots

Seja

u_{0} = 0

Calcular

u_{n + 1}

até que

Δ u_{n + 1} < 0.000001

u_{1} = 0.41421 \dots : Δ u_{1} = 0.41421 \dots

f (u_{0}) = 0^{5} - 0.41421 \dots \cdot 0^{4} + 1.17157 \dots \cdot 0^{3} - 16.48528 \dots \cdot 0^{2} + 5.82842 \dots \cdot 0 - 2.41421 \dots = - 2.41421 \dots

f^{^{'}} (u_{0}) = 5 \cdot 0^{4} - 1.65685 \dots \cdot 0^{3} + 3.51471 \dots \cdot 0^{2} - 32.97056 \dots \cdot 0 + 5.82842 \dots = 5.82842 \dots

u_{1} = 0.41421 \dots

Δ u_{1} = ∣ 0.41421 \dots - 0 ∣ = 0.41421 \dots

Δ u_{1} = 0.41421 \dots

u_{2} = 0.03272 \dots : Δ u_{2} = 0.38148 \dots

f (u_{1}) = 0.41421 \dots^{5} - 0.41421 \dots \cdot 0.41421 \dots^{4} + 1.17157 \dots \cdot 0.41421 \dots^{3} - 16.48528 \dots \cdot 0.41421 \dots^{2} + 5.82842 \dots \cdot 0.41421 \dots - 2.41421 \dots = - 2.74516 \dots

f^{^{'}} (u_{1}) = 5 \cdot 0.41421 \dots^{4} - 1.65685 \dots \cdot 0.41421 \dots^{3} + 3.51471 \dots \cdot 0.41421 \dots^{2} - 32.97056 \dots \cdot 0.41421 \dots + 5.82842 \dots = - 7.19595 \dots

u_{2} = 0.03272 \dots

Δ u_{2} = ∣ 0.03272 \dots - 0.41421 \dots ∣ = 0.38148 \dots

Δ u_{2} = 0.38148 \dots

u_{3} = 0.50422 \dots : Δ u_{3} = 0.47149 \dots

f (u_{2}) = 0.03272 \dots^{5} - 0.41421 \dots \cdot 0.03272 \dots^{4} + 1.17157 \dots \cdot 0.03272 \dots^{3} - 16.48528 \dots \cdot 0.03272 \dots^{2} + 5.82842 \dots \cdot 0.03272 \dots - 2.41421 \dots = - 2.24108 \dots

f^{^{'}} (u_{2}) = 5 \cdot 0.03272 \dots^{4} - 1.65685 \dots \cdot 0.03272 \dots^{3} + 3.51471 \dots \cdot 0.03272 \dots^{2} - 32.97056 \dots \cdot 0.03272 \dots + 5.82842 \dots = 4.75313 \dots

u_{3} = 0.50422 \dots

Δ u_{3} = ∣ 0.50422 \dots - 0.03272 \dots ∣ = 0.47149 \dots

Δ u_{3} = 0.47149 \dots

u_{4} = 0.14569 \dots : Δ u_{4} = 0.35852 \dots

f (u_{3}) = 0.50422 \dots^{5} - 0.41421 \dots \cdot 0.50422 \dots^{4} + 1.17157 \dots \cdot 0.50422 \dots^{3} - 16.48528 \dots \cdot 0.50422 \dots^{2} + 5.82842 \dots \cdot 0.50422 \dots - 2.41421 \dots = - 3.51061 \dots

f^{^{'}} (u_{3}) = 5 \cdot 0.50422 \dots^{4} - 1.65685 \dots \cdot 0.50422 \dots^{3} + 3.51471 \dots \cdot 0.50422 \dots^{2} - 32.97056 \dots \cdot 0.50422 \dots + 5.82842 \dots = - 9.79171 \dots

u_{4} = 0.14569 \dots

Δ u_{4} = ∣ 0.14569 \dots - 0.50422 \dots ∣ = 0.35852 \dots

Δ u_{4} = 0.35852 \dots

u_{5} = 1.88887 \dots : Δ u_{5} = 1.74318 \dots

f (u_{4}) = 0.14569 \dots^{5} - 0.41421 \dots \cdot 0.14569 \dots^{4} + 1.17157 \dots \cdot 0.14569 \dots^{3} - 16.48528 \dots \cdot 0.14569 \dots^{2} + 5.82842 \dots \cdot 0.14569 \dots - 2.41421 \dots = - 1.91147 \dots

f^{^{'}} (u_{4}) = 5 \cdot 0.14569 \dots^{4} - 1.65685 \dots \cdot 0.14569 \dots^{3} + 3.51471 \dots \cdot 0.14569 \dots^{2} - 32.97056 \dots \cdot 0.14569 \dots + 5.82842 \dots = 1.09654 \dots

u_{5} = 1.88887 \dots

Δ u_{5} = ∣ 1.88887 \dots - 0.14569 \dots ∣ = 1.74318 \dots

Δ u_{5} = 1.74318 \dots

u_{6} = 4.63637 \dots : Δ u_{6} = 2.74749 \dots

f (u_{5}) = 1.88887 \dots^{5} - 0.41421 \dots \cdot 1.88887 \dots^{4} + 1.17157 \dots \cdot 1.88887 \dots^{3} - 16.48528 \dots \cdot 1.88887 \dots^{2} + 5.82842 \dots \cdot 1.88887 \dots - 2.41421 \dots = - 23.55475 \dots

f^{^{'}} (u_{5}) = 5 \cdot 1.88887 \dots^{4} - 1.65685 \dots \cdot 1.88887 \dots^{3} + 3.51471 \dots \cdot 1.88887 \dots^{2} - 32.97056 \dots \cdot 1.88887 \dots + 5.82842 \dots = 8.57317 \dots

u_{6} = 4.63637 \dots

Δ u_{6} = ∣ 4.63637 \dots - 1.88887 \dots ∣ = 2.74749 \dots

Δ u_{6} = 2.74749 \dots

u_{7} = 3.79830 \dots : Δ u_{7} = 0.83806 \dots

f (u_{6}) = 4.63637 \dots^{5} - 0.41421 \dots \cdot 4.63637 \dots^{4} + 1.17157 \dots \cdot 4.63637 \dots^{3} - 16.48528 \dots \cdot 4.63637 \dots^{2} + 5.82842 \dots \cdot 4.63637 \dots - 2.41421 \dots = 1737.96023 \dots

f^{^{'}} (u_{6}) = 5 \cdot 4.63637 \dots^{4} - 1.65685 \dots \cdot 4.63637 \dots^{3} + 3.51471 \dots \cdot 4.63637 \dots^{2} - 32.97056 \dots \cdot 4.63637 \dots + 5.82842 \dots = 2073.76688 \dots

u_{7} = 3.79830 \dots

Δ u_{7} = ∣ 3.79830 \dots - 4.63637 \dots ∣ = 0.83806 \dots

Δ u_{7} = 0.83806 \dots

u_{8} = 3.17364 \dots : Δ u_{8} = 0.62465 \dots

f (u_{7}) = 3.79830 \dots^{5} - 0.41421 \dots \cdot 3.79830 \dots^{4} + 1.17157 \dots \cdot 3.79830 \dots^{3} - 16.48528 \dots \cdot 3.79830 \dots^{2} + 5.82842 \dots \cdot 3.79830 \dots - 2.41421 \dots = 550.45789 \dots

f^{^{'}} (u_{7}) = 5 \cdot 3.79830 \dots^{4} - 1.65685 \dots \cdot 3.79830 \dots^{3} + 3.51471 \dots \cdot 3.79830 \dots^{2} - 32.97056 \dots \cdot 3.79830 \dots + 5.82842 \dots = 881.21646 \dots

u_{8} = 3.17364 \dots

Δ u_{8} = ∣ 3.17364 \dots - 3.79830 \dots ∣ = 0.62465 \dots

Δ u_{8} = 0.62465 \dots

u_{9} = 2.74529 \dots : Δ u_{9} = 0.42835 \dots

f (u_{8}) = 3.17364 \dots^{5} - 0.41421 \dots \cdot 3.17364 \dots^{4} + 1.17157 \dots \cdot 3.17364 \dots^{3} - 16.48528 \dots \cdot 3.17364 \dots^{2} + 5.82842 \dots \cdot 3.17364 \dots - 2.41421 \dots = 167.42483 \dots

f^{^{'}} (u_{8}) = 5 \cdot 3.17364 \dots^{4} - 1.65685 \dots \cdot 3.17364 \dots^{3} + 3.51471 \dots \cdot 3.17364 \dots^{2} - 32.97056 \dots \cdot 3.17364 \dots + 5.82842 \dots = 390.85896 \dots

u_{9} = 2.74529 \dots

Δ u_{9} = ∣ 2.74529 \dots - 3.17364 \dots ∣ = 0.42835 \dots

Δ u_{9} = 0.42835 \dots

u_{10} = 2.50516 \dots : Δ u_{10} = 0.24012 \dots

f (u_{9}) = 2.74529 \dots^{5} - 0.41421 \dots \cdot 2.74529 \dots^{4} + 1.17157 \dots \cdot 2.74529 \dots^{3} - 16.48528 \dots \cdot 2.74529 \dots^{2} + 5.82842 \dots \cdot 2.74529 \dots - 2.41421 \dots = 45.99071 \dots

f^{^{'}} (u_{9}) = 5 \cdot 2.74529 \dots^{4} - 1.65685 \dots \cdot 2.74529 \dots^{3} + 3.51471 \dots \cdot 2.74529 \dots^{2} - 32.97056 \dots \cdot 2.74529 \dots + 5.82842 \dots = 191.52771 \dots

u_{10} = 2.50516 \dots

Δ u_{10} = ∣ 2.50516 \dots - 2.74529 \dots ∣ = 0.24012 \dots

Δ u_{10} = 0.24012 \dots

u_{11} = 2.42337 \dots : Δ u_{11} = 0.08179 \dots

f (u_{10}) = 2.50516 \dots^{5} - 0.41421 \dots \cdot 2.50516 \dots^{4} + 1.17157 \dots \cdot 2.50516 \dots^{3} - 16.48528 \dots \cdot 2.50516 \dots^{2} + 5.82842 \dots \cdot 2.50516 \dots - 2.41421 \dots = 9.50260 \dots

f^{^{'}} (u_{10}) = 5 \cdot 2.50516 \dots^{4} - 1.65685 \dots \cdot 2.50516 \dots^{3} + 3.51471 \dots \cdot 2.50516 \dots^{2} - 32.97056 \dots \cdot 2.50516 \dots + 5.82842 \dots = 116.17302 \dots

u_{11} = 2.42337 \dots

Δ u_{11} = ∣ 2.42337 \dots - 2.50516 \dots ∣ = 0.08179 \dots

Δ u_{11} = 0.08179 \dots

u_{12} = 2.41431 \dots : Δ u_{12} = 0.00905 \dots

f (u_{11}) = 2.42337 \dots^{5} - 0.41421 \dots \cdot 2.42337 \dots^{4} + 1.17157 \dots \cdot 2.42337 \dots^{3} - 16.48528 \dots \cdot 2.42337 \dots^{2} + 5.82842 \dots \cdot 2.42337 \dots - 2.41421 \dots = 0.86400 \dots

f^{^{'}} (u_{11}) = 5 \cdot 2.42337 \dots^{4} - 1.65685 \dots \cdot 2.42337 \dots^{3} + 3.51471 \dots \cdot 2.42337 \dots^{2} - 32.97056 \dots \cdot 2.42337 \dots + 5.82842 \dots = 95.43428 \dots

u_{12} = 2.41431 \dots

Δ u_{12} = ∣ 2.41431 \dots - 2.42337 \dots ∣ = 0.00905 \dots

Δ u_{12} = 0.00905 \dots

u_{13} = 2.41421 \dots : Δ u_{13} = 0.00010 \dots

f (u_{12}) = 2.41431 \dots^{5} - 0.41421 \dots \cdot 2.41431 \dots^{4} + 1.17157 \dots \cdot 2.41431 \dots^{3} - 16.48528 \dots \cdot 2.41431 \dots^{2} + 5.82842 \dots \cdot 2.41431 \dots - 2.41421 \dots = 0.00977 \dots

f^{^{'}} (u_{12}) = 5 \cdot 2.41431 \dots^{4} - 1.65685 \dots \cdot 2.41431 \dots^{3} + 3.51471 \dots \cdot 2.41431 \dots^{2} - 32.97056 \dots \cdot 2.41431 \dots + 5.82842 \dots = 93.27961 \dots

u_{13} = 2.41421 \dots

Δ u_{13} = ∣ 2.41421 \dots - 2.41431 \dots ∣ = 0.00010 \dots

Δ u_{13} = 0.00010 \dots

u_{14} = 2.41421 \dots : Δ u_{14} = 1.39211 E - 8

f (u_{13}) = 2.41421 \dots^{5} - 0.41421 \dots \cdot 2.41421 \dots^{4} + 1.17157 \dots \cdot 2.41421 \dots^{3} - 16.48528 \dots \cdot 2.41421 \dots^{2} + 5.82842 \dots \cdot 2.41421 \dots - 2.41421 \dots = 1.29821 E - 6

f^{^{'}} (u_{13}) = 5 \cdot 2.41421 \dots^{4} - 1.65685 \dots \cdot 2.41421 \dots^{3} + 3.51471 \dots \cdot 2.41421 \dots^{2} - 32.97056 \dots \cdot 2.41421 \dots + 5.82842 \dots = 93.25483 \dots

u_{14} = 2.41421 \dots

Δ u_{14} = ∣ 2.41421 \dots - 2.41421 \dots ∣ = 1.39211 E - 8

Δ u_{14} = 1.39211 E - 8

u \approx 2.41421 \dots

Aplicar a divisão longa

Eq u a t i o n 0 : \frac{u ^{5} - 0.41421 \dots u ^{4} + 1.17157 \dots u ^{3} - 16.48528 \dots u ^{2} + 5.82842 \dots u - 2.41421 \dots}{u - 2.41421 \dots} = u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1

u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1 \approx 0

Encontrar uma solução para

u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1 = 0

utilizando o método de Newton-Raphson:

Sem solução para

u \in R

u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1 = 0

Definição de método de Newton-Raphson

f (u) = u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1

Encontrar

f^{^{'}} (u) : 4 u^{3} + 6 u^{2} + 12 u - 2

\frac{d}{d u} (u^{4} + 2 u^{3} + 6 u^{2} - 2 u + 1)

Aplicar a regra da soma/diferença:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{4}) + \frac{d}{d u} (2 u^{3}) + \frac{d}{d u} (6 u^{2}) - \frac{d}{d u} (2 u) + \frac{d}{d u} (1)

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 u^{4 - 1}

Simplificar

= 4 u^{3}

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{3})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 3 u^{3 - 1}

Simplificar

= 6 u^{2}

\frac{d}{d u} (6 u^{2}) = 12 u

\frac{d}{d u} (6 u^{2})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d u} (u^{2})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 \cdot 2 u^{2 - 1}

Simplificar

= 12 u

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

Aplicar a regra da derivação:

\frac{d u}{d u} = 1

= 2 \cdot 1

Simplificar

= 2

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivada de uma constante:

\frac{d}{d x} (a) = 0

= 0

= 4 u^{3} + 6 u^{2} + 12 u - 2 + 0

Simplificar

= 4 u^{3} + 6 u^{2} + 12 u - 2

Seja

u_{0} = 1

Calcular

u_{n + 1}

até que

Δ u_{n + 1} < 0.000001

u_{1} = 0.6 : Δ u_{1} = 0.4

f (u_{0}) = 1^{4} + 2 \cdot 1^{3} + 6 \cdot 1^{2} - 2 \cdot 1 + 1 = 8

f^{^{'}} (u_{0}) = 4 \cdot 1^{3} + 6 \cdot 1^{2} + 12 \cdot 1 - 2 = 20

u_{1} = 0.6

Δ u_{1} = ∣ 0.6 - 1 ∣ = 0.4

Δ u_{1} = 0.4

u_{2} = 0.29338 \dots : Δ u_{2} = 0.30661 \dots

f (u_{1}) = 0. 6^{4} + 2 \cdot 0. 6^{3} + 6 \cdot 0. 6^{2} - 2 \cdot 0.6 + 1 = 2.5216

f^{^{'}} (u_{1}) = 4 \cdot 0. 6^{3} + 6 \cdot 0. 6^{2} + 12 \cdot 0.6 - 2 = 8.224

u_{2} = 0.29338 \dots

Δ u_{2} = ∣ 0.29338 \dots - 0.6 ∣ = 0.30661 \dots

Δ u_{2} = 0.30661 \dots

u_{3} = - 0.16852 \dots : Δ u_{3} = 0.46190 \dots

f (u_{2}) = 0.29338 \dots^{4} + 2 \cdot 0.29338 \dots^{3} + 6 \cdot 0.29338 \dots^{2} - 2 \cdot 0.29338 \dots + 1 = 0.98759 \dots

f^{^{'}} (u_{2}) = 4 \cdot 0.29338 \dots^{3} + 6 \cdot 0.29338 \dots^{2} + 12 \cdot 0.29338 \dots - 2 = 2.13808 \dots

u_{3} = - 0.16852 \dots

Δ u_{3} = ∣ - 0.16852 \dots - 0.29338 \dots ∣ = 0.46190 \dots

Δ u_{3} = 0.46190 \dots

u_{4} = 0.21863 \dots : Δ u_{4} = 0.38715 \dots

f (u_{3}) = (- 0.16852 \dots)^{4} + 2 (- 0.16852 \dots)^{3} + 6 (- 0.16852 \dots)^{2} - 2 (- 0.16852 \dots) + 1 = 1.49867 \dots

f^{^{'}} (u_{3}) = 4 (- 0.16852 \dots)^{3} + 6 (- 0.16852 \dots)^{2} + 12 (- 0.16852 \dots) - 2 = - 3.87099 \dots

u_{4} = 0.21863 \dots

Δ u_{4} = ∣ 0.21863 \dots - (- 0.16852 \dots) ∣ = 0.38715 \dots

Δ u_{4} = 0.38715 \dots

u_{5} = - 0.69789 \dots : Δ u_{5} = 0.91652 \dots

f (u_{4}) = 0.21863 \dots^{4} + 2 \cdot 0.21863 \dots^{3} + 6 \cdot 0.21863 \dots^{2} - 2 \cdot 0.21863 \dots + 1 = 0.87272 \dots

f^{^{'}} (u_{4}) = 4 \cdot 0.21863 \dots^{3} + 6 \cdot 0.21863 \dots^{2} + 12 \cdot 0.21863 \dots - 2 = 0.95220 \dots

u_{5} = - 0.69789 \dots

Δ u_{5} = ∣ - 0.69789 \dots - 0.21863 \dots ∣ = 0.91652 \dots

Δ u_{5} = 0.91652 \dots

u_{6} = - 0.14461 \dots : Δ u_{6} = 0.55328 \dots

f (u_{5}) = (- 0.69789 \dots)^{4} + 2 (- 0.69789 \dots)^{3} + 6 (- 0.69789 \dots)^{2} - 2 (- 0.69789 \dots) + 1 = 4.87554 \dots

f^{^{'}} (u_{5}) = 4 (- 0.69789 \dots)^{3} + 6 (- 0.69789 \dots)^{2} + 12 (- 0.69789 \dots) - 2 = - 8.81206 \dots

u_{6} = - 0.14461 \dots

Δ u_{6} = ∣ - 0.14461 \dots - (- 0.69789 \dots) ∣ = 0.55328 \dots

Δ u_{6} = 0.55328 \dots

u_{7} = 0.24442 \dots : Δ u_{7} = 0.38903 \dots

f (u_{6}) = (- 0.14461 \dots)^{4} + 2 (- 0.14461 \dots)^{3} + 6 (- 0.14461 \dots)^{2} - 2 (- 0.14461 \dots) + 1 = 1.40910 \dots

f^{^{'}} (u_{6}) = 4 (- 0.14461 \dots)^{3} + 6 (- 0.14461 \dots)^{2} + 12 (- 0.14461 \dots) - 2 = - 3.62201 \dots

u_{7} = 0.24442 \dots

Δ u_{7} = ∣ 0.24442 \dots - (- 0.14461 \dots) ∣ = 0.38903 \dots

Δ u_{7} = 0.38903 \dots

u_{8} = - 0.42403 \dots : Δ u_{8} = 0.66846 \dots

f (u_{7}) = 0.24442 \dots^{4} + 2 \cdot 0.24442 \dots^{3} + 6 \cdot 0.24442 \dots^{2} - 2 \cdot 0.24442 \dots + 1 = 0.90238 \dots

f^{^{'}} (u_{7}) = 4 \cdot 0.24442 \dots^{3} + 6 \cdot 0.24442 \dots^{2} + 12 \cdot 0.24442 \dots - 2 = 1.34994 \dots

u_{8} = - 0.42403 \dots

Δ u_{8} = ∣ - 0.42403 \dots - 0.24442 \dots ∣ = 0.66846 \dots

Δ u_{8} = 0.66846 \dots

u_{9} = 0.02045 \dots : Δ u_{9} = 0.44448 \dots

f (u_{8}) = (- 0.42403 \dots)^{4} + 2 (- 0.42403 \dots)^{3} + 6 (- 0.42403 \dots)^{2} - 2 (- 0.42403 \dots) + 1 = 2.80677 \dots

f^{^{'}} (u_{8}) = 4 (- 0.42403 \dots)^{3} + 6 (- 0.42403 \dots)^{2} + 12 (- 0.42403 \dots) - 2 = - 6.31459 \dots

u_{9} = 0.02045 \dots

Δ u_{9} = ∣ 0.02045 \dots - (- 0.42403 \dots) ∣ = 0.44448 \dots

Δ u_{9} = 0.44448 \dots

u_{10} = 0.56930 \dots : Δ u_{10} = 0.54885 \dots

f (u_{9}) = 0.02045 \dots^{4} + 2 \cdot 0.02045 \dots^{3} + 6 \cdot 0.02045 \dots^{2} - 2 \cdot 0.02045 \dots + 1 = 0.96162 \dots

f^{^{'}} (u_{9}) = 4 \cdot 0.02045 \dots^{3} + 6 \cdot 0.02045 \dots^{2} + 12 \cdot 0.02045 \dots - 2 = - 1.75205 \dots

u_{10} = 0.56930 \dots

Δ u_{10} = ∣ 0.56930 \dots - 0.02045 \dots ∣ = 0.54885 \dots

Δ u_{10} = 0.54885 \dots

Não se pode encontrar solução

As soluções são

u \approx - 0.41421 \dots, u \approx 2.41421 \dots

As soluções são

u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

Verifique soluções

Encontrar os pontos não definidos (singularidades):

u = 0

Tomar o(s) denominador(es) de

u^{4} + u^{- 4}

e comparar com zero

Resolver

u^{4} = 0 : u = 0

u^{4} = 0

Aplicar a regra

x^{n} = 0 \Rightarrow x = 0

u = 0

Tomar o(s) denominador(es) de

16 (u - u^{- 1}) + 2

e comparar com zero

u = 0

Os seguintes pontos são indefinidos

u = 0

Combinar os pontos indefinidos com as soluções:

u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

u = - 1, u = 1, u \approx - 0.41421 \dots, u \approx 2.41421 \dots

Substitua

u = e^{x},

solucione para

x

Resolver

e^{x} = - 1 :

Sem solução para

x \in R

e^{x} = - 1

a^{f (x)}

não pode ser zero ou negativa para

x \in R

Sem solu \overset{c}{\c} \tilde{a} o para x \in R

Resolver

e^{x} = 1 : x = 0

e^{x} = 1

Aplicar as propriedades dos expoentes

e^{x} = 1

f (x) = g (x)

, então

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1)

Aplicar as propriedades dos logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplificar

ln (1) : 0

ln (1)

Aplicar as propriedades dos logaritmos:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

Resolver

e^{x} = - 0.41421 \dots :

Sem solução para

x \in R

e^{x} = - 0.41421 \dots

a^{f (x)}

não pode ser zero ou negativa para

x \in R

Sem solu \overset{c}{\c} \tilde{a} o para x \in R

Resolver

e^{x} = 2.41421 \dots : x = ln (2.41421 \dots)

e^{x} = 2.41421 \dots

Aplicar as propriedades dos expoentes

e^{x} = 2.41421 \dots

f (x) = g (x)

, então

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (2.41421 \dots)

Aplicar as propriedades dos logaritmos:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2.41421 \dots)

x = ln (2.41421 \dots)

x = 0, x = ln (2.41421 \dots)

x = 0, x = ln (2.41421 \dots)

Gráfico

Exemplos populares

cos(θ)=((sqrt(3))/2)

cos (θ) = (\frac{3}{2})

cos(x+(3pi)/4)+cos(x-(3pi)/4)=1

cos (x + \frac{3 π}{4}) + cos (x - \frac{3 π}{4}) = 1

5-sin(θ)=cos(2θ)

5 - sin (θ) = cos (2 θ)

8sin(θ)+4sqrt(3)=0

8 sin (θ) + 43 = 0

solvefor x,sin(x)=(sqrt(3))/2

so l v e f or x, sin (x) = \frac{3}{2}