What is the tangent of f(x)= 1/(x^3),(-2,-1/8) ?

The tangent of f(x)= 1/(x^3),(-2,-1/8) is y=-3/16 x-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(x^3),(-2,-1/8)

tangent of

f (x) = \frac{1}{x ^{3}}, (- 2, - \frac{1}{8})

y = - \frac{3}{16} x - \frac{1}{2}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{3}} : \frac{df ( x )}{d x} = - \frac{3}{x ^{4}}

Compute the slope of

f (x) = \frac{1}{x ^{3}}

(- 2, - \frac{1}{8}) : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(- 2, - \frac{1}{8}) : y = - \frac{3}{16} x - \frac{1}{2}

y = - \frac{3}{16} x - \frac{1}{2}

\int x e^{\frac{x}{3}} d x

\frac{d y}{d x}, sin (x + y) = 3 x - 2 y

\int \frac{4}{4 x + 1} d x

e^{x} \frac{d y}{d x} = 2 x

(t - 1) cos (4 (t - 1)) + t^{2}

What is the tangent of f(x)= 1/(x^3),(-2,-1/8) ?
The tangent of f(x)= 1/(x^3),(-2,-1/8) is y=-3/16 x-1/2