What is the tangent of f(x)=4xe^x,\at x=0 ?

The tangent of f(x)=4xe^x,\at x=0 is y=4x

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tangent of f(x)=4xe^x,\at x=0

tangent of

f (x) = 4 x e^{x}, at x = 0

y = 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 4 x e^{x} : \frac{df ( x )}{d x} = 4 (e^{x} + e^{x} x)

Compute the slope of

f (x) = 4 x e^{x}

x = 0 : m = 4

Find the line with slope m=

4

and passing through

(0, 0) : y = 4 x

y = 4 x

\int \frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} d x

- \frac{6 x}{( x ^{2} - 4 ) ^{2}}

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

\frac{\partial}{\partial x} (x arctan (y))

n = 3 \sum \infty (\frac{9}{11})^{- n}

What is the tangent of f(x)=4xe^x,\at x=0 ?
The tangent of f(x)=4xe^x,\at x=0 is y=4x