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

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

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

tangent of

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

y = 2 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

\frac{\partial}{\partial x} (4 ln (x y) + 3 ln (yz) + ln (x z))

x = y^{2} - 4 y, x = y - y^{2}

2 y^{^{''}} - 15 y^{^{'}} - 8 y = 0

\int_{- 1}^{1} x - 1 - x^{2} d x

y = (x^{6} - 10) x

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