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tangent f(x)=(sqrt(x))/(4x-7),\at x=1

解答

切线

f (x) = \frac{x}{4 x - 7}, at x = 1

解答

y = - \frac{11}{18} x + \frac{5}{18}

求解步骤

找到切点:

(1, - \frac{1}{3})

计算

f (x) = \frac{x}{4 x - 7}

的斜率:

\frac{df ( x )}{d x} = \frac{- 4 x - 7}{2 x ( 4 x - 7 ) ^{2}}

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 = - \frac{11}{18}

找到斜率 m=

- \frac{11}{18}

且穿过

(1, - \frac{1}{3})

的直线:

y = - \frac{11}{18} x + \frac{5}{18}

y = - \frac{11}{18} x + \frac{5}{18}

作图

流行的例子

tangent-5x^2+x-2

t an g e n t - 5 x^{2} + x - 2

integral from 0 to 9 of 1/(81+x^2)

\int_{0}^{9} \frac{1}{81 + x ^{2}} d x

y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 5 y = - 20 e^{t}, y (0) = - 4, y^{^{'}} (0) = 1

integral from 1 to 4 of 6e^{sqrt(x)}

\int_{1}^{4} 6 e^{x} d x

critical f(x,y)=(x)^2+(y)^4

cr i t i c a l f (x, y) = (x)^{2} + (y)^{4}