What is the extreme g(x)=e^{-x^4},-3<= x<= 1 ?

The extreme g(x)=e^{-x^4},-3<= x<= 1 is Minimum(-3, 1/(e^{81)}),Maximum(0,1),Minimum(1, 1/e)

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extreme g(x)=e^{-x^4},-3<= x<= 1

extreme points

g (x) = e^{- x^{4}}, - 3 \leq x \leq 1

Minimum (- 3, \frac{1}{e ^{81}}), Maximum (0, 1), Minimum (1, \frac{1}{e})

Solution steps

Find the critical points:

x = - 3, x = 0, x = 1

Domain of

e^{- x^{4}}, - 3 \leq x \leq 1 : - 3 \leq x \leq 1

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 3

into

e^{- x^{4}} \Rightarrow y = \frac{1}{e ^{81}}

Minimum (- 3, \frac{1}{e ^{81}})

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

Plug the extreme point

x = 1

into

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

Minimum (1, \frac{1}{e})

Minimum (- 3, \frac{1}{e ^{81}}), Maximum (0, 1), Minimum (1, \frac{1}{e})

f (x, y) = x^{2} + y^{2} - 4 x + 3 y

f (x, y) = 2^{3} - y^{2} + 2 x y + 1

y = x^{2} + 2

f (x) = 2^{(x + 0.528)} - 10 (x + 0.528)

y = x^{2} - 5

What is the extreme g(x)=e^{-x^4},-3<= x<= 1 ?
The extreme g(x)=e^{-x^4},-3<= x<= 1 is Minimum(-3, 1/(e^{81)}),Maximum(0,1),Minimum(1, 1/e)