What is the extreme f(x)=e^{x^2-4},(-2,2) ?

The extreme f(x)=e^{x^2-4},(-2,2) is Minimum(0, 1/(e^4))

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extreme f(x)=e^{x^2-4},(-2,2)

extreme

Minimum (0, \frac{1}{e ^{4}})

Solution steps

Find the critical points:

x = 0

Domain of

e^{x^{2} - 4}, (- 2, 2) : - 2 < x < 2

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = 0

into

e^{x^{2} - 4} \Rightarrow y = \frac{1}{e ^{4}}

Minimum (0, \frac{1}{e ^{4}})

What is the extreme f(x)=e^{x^2-4},(-2,2) ?
The extreme f(x)=e^{x^2-4},(-2,2) is Minimum(0, 1/(e^4))