What is the tangent of f(x)=2-xe^x,\at x=0 ?

The tangent of f(x)=2-xe^x,\at x=0 is y=-x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2-xe^x,\at x=0

tangent of

f (x) = 2 - x e^{x}, at x = 0

y = - x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = 2 - x e^{x} : \frac{df ( x )}{d x} = - e^{x} x - 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 = - 1

Find the line with slope m=

- 1

and passing through

(0, 2) : y = - x + 2

y = - x + 2

y^{^{'}} = e^{- y} (2 x - 4), y (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 t}

y = ln (x^{3} e^{2 x})

\frac{x ^{3} + 2}{3}

(4 + x) \frac{d y}{d x} = y^{3}

What is the tangent of f(x)=2-xe^x,\at x=0 ?
The tangent of f(x)=2-xe^x,\at x=0 is y=-x+2