What is the tangent of f(x)=6+sqrt(1-x),\at x=-8 ?

The tangent of f(x)=6+sqrt(1-x),\at x=-8 is y=-1/6 x+23/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=6+sqrt(1-x),\at x=-8

tangent of

f (x) = 6 + 1 - x, at x = - 8

y = - \frac{1}{6} x + \frac{23}{3}

Solution steps

Find the tangent point:

(- 8, 9)

Find the slope of

f (x) = 6 + 1 - x : \frac{df ( x )}{d x} = - \frac{1}{2 - x + 1}

Compute the slope of

f (x) = 6 + 1 - x

x = - 8 : m = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(- 8, 9) : y = - \frac{1}{6} x + \frac{23}{3}

y = - \frac{1}{6} x + \frac{23}{3}

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

\frac{d}{d t} (e^{- 4 t})

x \frac{d y}{d x} + 3 x^{2} + 4 y = 0

x \cdot e^{x}

x (x^{3} + 8)

What is the tangent of f(x)=6+sqrt(1-x),\at x=-8 ?
The tangent of f(x)=6+sqrt(1-x),\at x=-8 is y=-1/6 x+23/3