What is the tangent of f(x)=x^2-x^4,\at x=1 ?

The tangent of f(x)=x^2-x^4,\at x=1 is y=-2x+2

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tangent of f(x)=x^2-x^4,\at x=1

tangent of

f (x) = x^{2} - x^{4}, at x = 1

y = - 2 x + 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{2} - x^{4} : \frac{df ( x )}{d x} = 2 x - 4 x^{3}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

\frac{\partial}{\partial x} (9 x^{3 y})

\int \frac{sin ( ln ( 45 x ) )}{x} d x

\int \frac{1}{k ( ln ( k ) ) ^{2}} d k

x \to 0 lim (\frac{1}{sin ( \frac{1}{x} )})

\frac{d}{d x} ((2 x - 1)^{4})

What is the tangent of f(x)=x^2-x^4,\at x=1 ?
The tangent of f(x)=x^2-x^4,\at x=1 is y=-2x+2