What is the extreme f(x)=x^2e^x[-3.1] ?

The extreme f(x)=x^2e^x[-3.1] is Minimum(-2,-1.67815…),Maximum(0,0)

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extreme f(x)=x^2e^x[-3.1]

extreme points

f (x) = x^{2} e^{x} [- 3.1]

Minimum (- 2, - 1.67815 \dots), Maximum (0, 0)

Solution steps

Find the critical points:

x = - 2, x = 0

Domain of

x^{2} e^{x} [- 3.1] : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug the extreme point

x = - 2

into

x^{2} e^{x} [- 3.1] \Rightarrow y = - 1.67815 \dots

Minimum (- 2, - 1.67815 \dots)

Plug the extreme point

x = 0

into

x^{2} e^{x} [- 3.1] \Rightarrow y = 0

Maximum (0, 0)

Minimum (- 2, - 1.67815 \dots), Maximum (0, 0)

\frac{1}{x ^{2} - 1}

f (x) = x^{3} - 5 x^{2} + 3 x + 12

f (x) = (x - 8) e^{- 9 x}

3 x^{4} - 8 x^{3} + 6 x^{2}

\frac{1}{x ^{2} + 1}

What is the extreme f(x)=x^2e^x[-3.1] ?
The extreme f(x)=x^2e^x[-3.1] is Minimum(-2,-1.67815…),Maximum(0,0)