What is the extreme f(x)=-2/5 x^{5/2}+32x^{1/2} ?

The extreme f(x)=-2/5 x^{5/2}+32x^{1/2} is Minimum(0,0),Maximum(4, 256/5)

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extreme f(x)=-2/5 x^{5/2}+32x^{1/2}

extreme points

f (x) = - \frac{2}{5} x^{\frac{5}{2}} + 32 x^{\frac{1}{2}}

Minimum (0, 0), Maximum (4, \frac{256}{5})

Solution steps

f^{'} (x) = \frac{16 - x ^{2}}{x ^{\frac{1}{2}}}

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

- \frac{2}{5} x^{\frac{5}{2}} + 32 x^{\frac{1}{2}} : 0

Minimum (0, 0)

Plug

x = 4

into

- \frac{2}{5} x^{\frac{5}{2}} + 32 x^{\frac{1}{2}} : \frac{256}{5}

Maximum (4, \frac{256}{5})

Minimum (0, 0), Maximum (4, \frac{256}{5})

f (x) = (\frac{- 2 x}{x ^{2} + 7})

f (x, y) = (x^{2} - 8 x) (y^{2} - 3 y)

F (x) = \frac{5 x ^{2}}{x ^{2} - 16}

f (x) = \frac{I n ( x ^{2} - 1 )}{x ^{2} - 1}

16 (- \frac{1}{x ^{2}} - \frac{2}{x ^{3}})

What is the extreme f(x)=-2/5 x^{5/2}+32x^{1/2} ?
The extreme f(x)=-2/5 x^{5/2}+32x^{1/2} is Minimum(0,0),Maximum(4, 256/5)