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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

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

Maximum (- 1, e)

Plug the extreme point

x = 0

into

x^{2} e^{- x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

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

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

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

f (x, y) = 36 - 9 x^{2} - 4 y^{2}

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

f (x, y) = (x^{2} + y^{2}) \cdot e^{- x - y}

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

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