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

The extreme f(x)=(x^{5/3})/(2+x) is Minimum(-5,(5(-5)^{2/3})/3),Saddle(0,0)

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

extreme points

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

Minimum (- 5, \frac{5 ( - 5 ) ^{\frac{2}{3}}}{3}), Saddle (0, 0)

Solution steps

f^{'} (x) = \frac{10 x ^{\frac{2}{3}} + 2 x ^{\frac{5}{3}}}{3 ( 2 + x ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 5

into

\frac{x ^{\frac{5}{3}}}{2 + x} : \frac{5 ( - 5 ) ^{\frac{2}{3}}}{3}

Minimum (- 5, \frac{5 ( - 5 ) ^{\frac{2}{3}}}{3})

Plug

x = 0

into

\frac{x ^{\frac{5}{3}}}{2 + x} : 0

Saddle (0, 0)

Minimum (- 5, \frac{5 ( - 5 ) ^{\frac{2}{3}}}{3}), Saddle (0, 0)

f (x) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

f (x, y) = \frac{23}{2} y^{4} - 46 x y + 23 x^{2}

- 8 x^{\frac{5}{3}} + 3 x

f (x, y) = 2 x^{3} + 6 x y + 3 y^{2}

f (x, y) = 16 x 3 y - 2 x^{2} - 4 x - 16 y + 2

What is the extreme f(x)=(x^{5/3})/(2+x) ?
The extreme f(x)=(x^{5/3})/(2+x) is Minimum(-5,(5(-5)^{2/3})/3),Saddle(0,0)