What is the tangent of f(x)=-1/(2sqrt(1-x)),\at x=-8 ?

The tangent of f(x)=-1/(2sqrt(1-x)),\at x=-8 is y=-1/108 x-13/54

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=-1/(2sqrt(1-x)),\at x=-8

tangent of

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

y = - \frac{1}{108} x - \frac{13}{54}

Solution steps

Find the tangent point:

(- 8, - \frac{1}{6})

Find the slope of

f (x) = - \frac{1}{2 1 - x} : \frac{df ( x )}{d x} = - \frac{1}{4 ( 1 - x ) ^{\frac{3}{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{1}{108}

Find the line with slope m=

- \frac{1}{108}

and passing through

(- 8, - \frac{1}{6}) : y = - \frac{1}{108} x - \frac{13}{54}

y = - \frac{1}{108} x - \frac{13}{54}

What is the tangent of f(x)=-1/(2sqrt(1-x)),\at x=-8 ?
The tangent of f(x)=-1/(2sqrt(1-x)),\at x=-8 is y=-1/108 x-13/54