What is the tangent of f(x)=8(e^x+e^xx),\at x=0 ?

The tangent of f(x)=8(e^x+e^xx),\at x=0 is y=16x+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=8(e^x+e^xx),\at x=0

tangent of

f (x) = 8 (e^{x} + e^{x} x), at x = 0

y = 16 x + 8

Solution steps

Find the tangent point:

(0, 8)

Find the slope of

f (x) = 8 (e^{x} + e^{x} x) : \frac{df ( x )}{d x} = 8 (e^{x} x + 2 e^{x})

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 = 16

Find the line with slope m=

16

and passing through

(0, 8) : y = 16 x + 8

y = 16 x + 8

y^{^{'}} = \frac{x}{( 1 + 2 y )}

\frac{d ^{3}}{d x ^{3}} (3 t^{3} - \frac{2}{t ^{2}})

\int (x^{3} + x + \frac{1}{x ^{3}}) d x

\int_{0}^{\frac{π}{2}} - 10 sin^{2} (3 x) d x

\int \frac{x}{( 1 + 2 x )} d x

What is the tangent of f(x)=8(e^x+e^xx),\at x=0 ?
The tangent of f(x)=8(e^x+e^xx),\at x=0 is y=16x+8