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 points

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

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}})

f (x, y) = \frac{y}{x} + x^{2} y + 6 x

y = - \frac{3}{2 x + 2}

4 sin (x) + 5 cos (x)

f (x) = - 3 x^{2} + 24 x - 5

f (x) = x^{2} + y^{2} - 9

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))