What is the extreme f(x)=e^{-x^6},-2<= x<= 1 ?

The extreme f(x)=e^{-x^6},-2<= x<= 1 is Minimum(-2, 1/(e^{64)}),Maximum(0,1),Minimum(1, 1/e)

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extreme f(x)=e^{-x^6},-2<= x<= 1

extreme points

f (x) = e^{- x^{6}}, - 2 \leq x \leq 1

Minimum (- 2, \frac{1}{e ^{64}}), Maximum (0, 1), Minimum (1, \frac{1}{e})

Solution steps

Find the critical points:

x = - 2, x = 0, x = 1

Domain of

e^{- x^{6}}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

e^{- x^{6}} \Rightarrow y = \frac{1}{e ^{64}}

Minimum (- 2, \frac{1}{e ^{64}})

Plug the extreme point

x = 0

into

e^{- x^{6}} \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 1

into

e^{- x^{6}} \Rightarrow y = \frac{1}{e}

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

Minimum (- 2, \frac{1}{e ^{64}}), Maximum (0, 1), Minimum (1, \frac{1}{e})

f (x) = x^{6} - 3 x^{4}

f (x, y) = 4 e^{- 5 x - 5 y}

f (x, y) = x^{2} + 3 x y + 4 y^{2} - 6 x + 2 y

f (x) = 250 + 8 x^{3} + x^{4}

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

What is the extreme f(x)=e^{-x^6},-2<= x<= 1 ?
The extreme f(x)=e^{-x^6},-2<= x<= 1 is Minimum(-2, 1/(e^{64)}),Maximum(0,1),Minimum(1, 1/e)