What is the extreme f(x)=e^{x^2-9},-3<= x<= 3 ?

The extreme f(x)=e^{x^2-9},-3<= x<= 3 is Maximum(-3,1),Minimum(0, 1/(e^9)),Maximum(3,1)

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extreme f(x)=e^{x^2-9},-3<= x<= 3

extreme

Maximum (- 3, 1), Minimum (0, \frac{1}{e ^{9}}), Maximum (3, 1)

Solution steps

Find the critical points:

x = - 3, x = 0, x = 3

Domain of

e^{x^{2} - 9}, - 3 \leq x \leq 3 : - 3 \leq x \leq 3

Find intervals:

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 3

Plug the extreme point

x = - 3

into

e^{x^{2} - 9} \Rightarrow y = 1

Maximum (- 3, 1)

Plug the extreme point

x = 0

into

e^{x^{2} - 9} \Rightarrow y = \frac{1}{e ^{9}}

Minimum (0, \frac{1}{e ^{9}})

Plug the extreme point

x = 3

into

e^{x^{2} - 9} \Rightarrow y = 1

Maximum (3, 1)

Maximum (- 3, 1), Minimum (0, \frac{1}{e ^{9}}), Maximum (3, 1)

What is the extreme f(x)=e^{x^2-9},-3<= x<= 3 ?
The extreme f(x)=e^{x^2-9},-3<= x<= 3 is Maximum(-3,1),Minimum(0, 1/(e^9)),Maximum(3,1)