What is the extreme f(x)=(x^{5/2-1}*e^{-x/2}) ?

The extreme f(x)=(x^{5/2-1}*e^{-x/2}) is Minimum(0,0),Maximum(3,(3*3^{1/2})/(e^{3/2)})

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extreme f(x)=(x^{5/2-1}*e^{-x/2})

extreme points

f (x) = (x^{\frac{5}{2} - 1} \cdot e^{- \frac{x}{2}})

Minimum (0, 0), Maximum (3, \frac{3 \cdot 3 ^{\frac{1}{2}}}{e ^{\frac{3}{2}}})

Solution steps

f^{'} (x) = \frac{e ^{- \frac{x}{2}} x ^{\frac{1}{2}} ( - x + 3 )}{2}

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

x^{\frac{5}{2} - 1} \cdot e^{- \frac{x}{2}} : 0

Minimum (0, 0)

Plug

x = 3

into

x^{\frac{5}{2} - 1} \cdot e^{- \frac{x}{2}} : \frac{3 \cdot 3 ^{\frac{1}{2}}}{e ^{\frac{3}{2}}}

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

Minimum (0, 0), Maximum (3, \frac{3 \cdot 3 ^{\frac{1}{2}}}{e ^{\frac{3}{2}}})

What is the extreme f(x)=(x^{5/2-1}*e^{-x/2}) ?
The extreme f(x)=(x^{5/2-1}*e^{-x/2}) is Minimum(0,0),Maximum(3,(3*3^{1/2})/(e^{3/2)})