What is the extreme f(x)=e^{x^2}(x^2-5)+8 ?

The extreme f(x)=e^{x^2}(x^2-5)+8 is Minimum(-2,-e^4+8),Maximum(0,3),Minimum(2,-e^4+8)

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extreme f(x)=e^{x^2}(x^2-5)+8

extreme points

f (x) = e^{x^{2}} (x^{2} - 5) + 8

Minimum (- 2, - e^{4} + 8), Maximum (0, 3), Minimum (2, - e^{4} + 8)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

e^{x^{2}} (x^{2} - 5) + 8 : - e^{4} + 8

Minimum (- 2, - e^{4} + 8)

Plug

x = 0

into

e^{x^{2}} (x^{2} - 5) + 8 : 3

Maximum (0, 3)

Plug

x = 2

into

e^{x^{2}} (x^{2} - 5) + 8 : - e^{4} + 8

Minimum (2, - e^{4} + 8)

Minimum (- 2, - e^{4} + 8), Maximum (0, 3), Minimum (2, - e^{4} + 8)

f (x) = 2 + 5 e^{- 0.1 x}

\frac{3 x}{5 - x}

f (x) = x^{3} + 9 x^{2} + 15 x + 1

- 0.001 x^{2} + 4.6 x - 90

f (x) = x - 4 \cdot ln (x - 4)

What is the extreme f(x)=e^{x^2}(x^2-5)+8 ?
The extreme f(x)=e^{x^2}(x^2-5)+8 is Minimum(-2,-e^4+8),Maximum(0,3),Minimum(2,-e^4+8)