What is the extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1 ?

The extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1 is Maximum(0,-1),Minimum(1,-19/15)

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extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1

extreme

Maximum (0, - 1), Minimum (1, - \frac{19}{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} - 1 : - 1

Maximum (0, - 1)

Plug

x = 1

into

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 1 : - \frac{19}{15}

Minimum (1, - \frac{19}{15})

Maximum (0, - 1), Minimum (1, - \frac{19}{15})

What is the extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1 ?
The extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1 is Maximum(0,-1),Minimum(1,-19/15)