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}}})

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

f (x, y) = 242 y^{2} + x^{2} - x^{2} y

- 4 x + 9 x^{\frac{4}{9}}

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

f (x, y) = \frac{1}{3} x^{3} + y^{3} + 2 x^{2} - 12 x - 3 y

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)})