What is the extreme f(x)=x^2e^{4x} ?

The extreme f(x)=x^2e^{4x} is Maximum(-1/2 , 1/(4e^2)),Minimum(0,0)

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extreme f(x)=x^2e^{4x}

extreme points

f (x) = x^{2} e^{4 x}

Maximum (- \frac{1}{2}, \frac{1}{4 e ^{2}}), Minimum (0, 0)

Solution steps

f^{'} (x) = 2 x e^{4 x} + e^{4 x} \cdot 4 x^{2}

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{2}

into

x^{2} e^{4 x} : \frac{1}{4 e ^{2}}

Maximum (- \frac{1}{2}, \frac{1}{4 e ^{2}})

Plug

x = 0

into

x^{2} e^{4 x} : 0

Minimum (0, 0)

Maximum (- \frac{1}{2}, \frac{1}{4 e ^{2}}), Minimum (0, 0)

f (x) = (x + 1)^{5} - 5 x - 2

f (x) = 5 x^{5} - 3 x^{3}

f (x) = \frac{- 5 x ^{2} + 16 x}{2 4 - x}

f (x) = x^{3} - 3 x^{2} - 3 x - 1

f (x) = 4 x^{2} - 8 x + 3

What is the extreme f(x)=x^2e^{4x} ?
The extreme f(x)=x^2e^{4x} is Maximum(-1/2 , 1/(4e^2)),Minimum(0,0)