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

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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}

\int \frac{1}{u 7 - u ^{2}} d u

f^{^{'}} (x) = \frac{8 x ^{5} - 6 x ^{9}}{x ^{4}}

\frac{d}{d x} (ln (1 + e^{2 x}))

x \to - 1 lim (\frac{x ^{2} - x}{x})

\int_{2}^{\infty} \frac{1}{x ^{2} - 1} d x

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