What is the tangent of f(x)=9+8x+6xe^x,\at x=0 ?

The tangent of f(x)=9+8x+6xe^x,\at x=0 is y=14x+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=9+8x+6xe^x,\at x=0

tangent of

f (x) = 9 + 8 x + 6 x e^{x}, at x = 0

y = 14 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

f (x) = 9 + 8 x + 6 x e^{x} : \frac{df ( x )}{d x} = 6 (e^{x} x + e^{x}) + 8

Compute the slope of

f (x) = 9 + 8 x + 6 x e^{x}

x = 0 : m = 14

Find the line with slope m=

14

and passing through

(0, 9) : y = 14 x + 9

y = 14 x + 9

\frac{\partial}{\partial x} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

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

n = 1 \sum \infty 3 (5)^{- n}

x \to 0 lim (9 x - 8 x)

\int_{0}^{\frac{1}{4}} arcsin (2 x) d x

What is the tangent of f(x)=9+8x+6xe^x,\at x=0 ?
The tangent of f(x)=9+8x+6xe^x,\at x=0 is y=14x+9