What is the tangent of f(x)=7xe^x,\at x=0 ?

The tangent of f(x)=7xe^x,\at x=0 is y=7x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=7xe^x,\at x=0

tangent of

f (x) = 7 x e^{x}, at x = 0

y = 7 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 7 x e^{x} : \frac{df ( x )}{d x} = 7 (e^{x} + e^{x} x)

Compute the slope of

f (x) = 7 x e^{x}

x = 0 : m = 7

Find the line with slope m=

7

and passing through

(0, 0) : y = 7 x

y = 7 x

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

y = \frac{6 + x f ( x )}{x}

\int y^{2} 2 y^{3} + 1 d y

\frac{d}{d x} (\frac{x ^{2} + x + 1}{x ^{2} - x + 1})

\frac{( x ^{2} )}{2}

What is the tangent of f(x)=7xe^x,\at x=0 ?
The tangent of f(x)=7xe^x,\at x=0 is y=7x