What is the extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-3 ?

The extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-3 is Maximum(0,-3),Minimum(1,-49/15)

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extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-3

extreme points

f (x) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 3

Maximum (0, - 3), Minimum (1, - \frac{49}{15})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 3 : - 3

Maximum (0, - 3)

Plug

x = 1

into

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 3 : - \frac{49}{15}

Minimum (1, - \frac{49}{15})

Maximum (0, - 3), Minimum (1, - \frac{49}{15})

f (x) = 16 x + \frac{9}{x}

f (x) = x + \frac{22}{x}

y = 6 x - 7 x^{\frac{6}{7}}

f (x, y) = ln (x^{2} + y^{2} - 64)

(x + 6)^{\frac{2}{3}}

What is the extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-3 ?
The extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-3 is Maximum(0,-3),Minimum(1,-49/15)