What is the extreme f(x)=x^2e^{16x} ?

The extreme f(x)=x^2e^{16x} is Maximum(-1/8 , 1/(64e^2)),Minimum(0,0)

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extreme f(x)=x^2e^{16x}

extreme points

f (x) = x^{2} e^{16 x}

Maximum (- \frac{1}{8}, \frac{1}{64 e ^{2}}), Minimum (0, 0)

Solution steps

f^{'} (x) = 2 x e^{16 x} + e^{16 x} \cdot 16 x^{2}

Find intervals:

Decreasing

: - \frac{1}{8} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{8}

into

x^{2} e^{16 x} : \frac{1}{64 e ^{2}}

Maximum (- \frac{1}{8}, \frac{1}{64 e ^{2}})

Plug

x = 0

into

x^{2} e^{16 x} : 0

Minimum (0, 0)

Maximum (- \frac{1}{8}, \frac{1}{64 e ^{2}}), Minimum (0, 0)

f (x) = 3 t^{3} + 2 t^{2} - 10 t + 600

3 x^{2} - x - 2

f (x) = \frac{( x ^{2} - 11 )}{x - 6}

f (x, y) = x^{2} y + y^{3} - 27 y

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

What is the extreme f(x)=x^2e^{16x} ?
The extreme f(x)=x^2e^{16x} is Maximum(-1/8 , 1/(64e^2)),Minimum(0,0)