What is the tangent of f(x)=-1-4e^{-x/3},\at x=8 ?

The tangent of f(x)=-1-4e^{-x/3},\at x=8 is y= 4/(3e^{8/3)}x-(44)/(3e^{8/3)}-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=-1-4e^{-x/3},\at x=8

tangent of

f (x) = - 1 - 4 e^{- \frac{x}{3}}, at x = 8

y = \frac{4}{3 e ^{\frac{8}{3}}} x - \frac{44}{3 e ^{\frac{8}{3}}} - 1

Solution steps

Find the tangent point:

(8, - 1 - \frac{4}{e ^{\frac{8}{3}}})

Find the slope of

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

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{4}{3 e ^{\frac{8}{3}}}

Find the line with slope m=

\frac{4}{3 e ^{\frac{8}{3}}}

and passing through

(8, - 1 - \frac{4}{e ^{\frac{8}{3}}}) : y = \frac{4}{3 e ^{\frac{8}{3}}} x - \frac{44}{3 e ^{\frac{8}{3}}} - 1

y = \frac{4}{3 e ^{\frac{8}{3}}} x - \frac{44}{3 e ^{\frac{8}{3}}} - 1

x y^{^{'}} - y = 2 x^{2} - x - 1

\int sin^{2} (\frac{π x}{l}) d x

\frac{d}{d x} (2 (x + 2)^{\frac{3}{2}})

\int \frac{9}{1 + x ^{2}} d x

\int (\frac{5}{x} - \frac{3}{x ^{4}}) d x

What is the tangent of f(x)=-1-4e^{-x/3},\at x=8 ?
The tangent of f(x)=-1-4e^{-x/3},\at x=8 is y= 4/(3e^{8/3)}x-(44)/(3e^{8/3)}-1