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

The tangent of f(x)=8e^x+x,\at x=8 is y=(8e^8+1)x-56e^8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = (8 e^{8} + 1) x - 56 e^{8}

Solution steps

Find the tangent point:

(8, 8 e^{8} + 8)

Find the slope of

f (x) = 8 e^{x} + x : \frac{df ( x )}{d x} = 8 e^{x} + 1

Compute the slope of

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

x = 8 : m = 8 e^{8} + 1

Find the line with slope m=

8 e^{8} + 1

and passing through

(8, 8 e^{8} + 8) : y = (8 e^{8} + 1) x - 56 e^{8}

y = (8 e^{8} + 1) x - 56 e^{8}

x \to \infty lim (- 1 + e^{(x + 1)})

x \to \infty lim ((6)^{x})

\int \frac{x ^{4} e ^{x} - 4 x ^{3}}{x ^{4}} d x

y^{^{'}} = 9.8 - 0.2 y, y (0) = 100

n = - 4 \sum \infty (- \frac{4}{9})^{n}

What is the tangent of f(x)=8e^x+x,\at x=8 ?
The tangent of f(x)=8e^x+x,\at x=8 is y=(8e^8+1)x-56e^8