What is the extreme f(x)=x^2e^x-7 ?

The extreme f(x)=x^2e^x-7 is Maximum(-2, 4/(e^2)-7),Minimum(0,-7)

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extreme f(x)=x^2e^x-7

extreme points

f (x) = x^{2} e^{x} - 7

Maximum (- 2, \frac{4}{e ^{2}} - 7), Minimum (0, - 7)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

x^{2} e^{x} - 7 : \frac{4}{e ^{2}} - 7

Maximum (- 2, \frac{4}{e ^{2}} - 7)

Plug

x = 0

into

x^{2} e^{x} - 7 : - 7

Minimum (0, - 7)

Maximum (- 2, \frac{4}{e ^{2}} - 7), Minimum (0, - 7)

f (x) = \frac{2 x - 7}{x ^{2} - 5 x + 6}

16 x + 13 y = 26

x^{3} + 1

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

- 16 x^{2} + 20

What is the extreme f(x)=x^2e^x-7 ?
The extreme f(x)=x^2e^x-7 is Maximum(-2, 4/(e^2)-7),Minimum(0,-7)