What is the extreme f(x)=xe^{x/2},-3<= x<= 1 ?

The extreme f(x)=xe^{x/2},-3<= x<= 1 is Maximum(-3,-3/(e^{3/2)}),Minimum(-2,-2/e),Maximum(1,e^{1/2})

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extreme f(x)=xe^{x/2},-3<= x<= 1

extreme points

f (x) = x e^{\frac{x}{2}}, - 3 \leq x \leq 1

Maximum (- 3, - \frac{3}{e ^{\frac{3}{2}}}), Minimum (- 2, - \frac{2}{e}), Maximum (1, e^{\frac{1}{2}})

Solution steps

Find the critical points:

x = - 3, x = - 2, x = 1

Domain of

x e^{\frac{x}{2}}, - 3 \leq x \leq 1 : - 3 \leq x \leq 1

Find intervals:

Decreasing

: - 3 < x < - 2,

Increasing

: - 2 < x < 1

Plug the extreme point

x = - 3

into

x e^{\frac{x}{2}} \Rightarrow y = - \frac{3}{e ^{\frac{3}{2}}}

Maximum (- 3, - \frac{3}{e ^{\frac{3}{2}}})

Plug the extreme point

x = - 2

into

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

Minimum (- 2, - \frac{2}{e})

Plug the extreme point

x = 1

into

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

Maximum (1, e^{\frac{1}{2}})

Maximum (- 3, - \frac{3}{e ^{\frac{3}{2}}}), Minimum (- 2, - \frac{2}{e}), Maximum (1, e^{\frac{1}{2}})

x + y - x y

p (x) = 3 x^{4} + 2 x^{3} + 5 x^{2} + a x - 2

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

f (x) = \frac{x ^{2}}{9} + 11 + \frac{6}{x}

f (x) = - 12 x - 6

What is the extreme f(x)=xe^{x/2},-3<= x<= 1 ?
The extreme f(x)=xe^{x/2},-3<= x<= 1 is Maximum(-3,-3/(e^{3/2)}),Minimum(-2,-2/e),Maximum(1,e^{1/2})