What is the extreme f(x)=x^{(2/5)}(7/(2+x)) ?

The extreme f(x)=x^{(2/5)}(7/(2+x)) is Minimum(0,0),Maximum(4/3 ,(7*4^{2/5}*3^{3/5})/(10))

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extreme f(x)=x^{(2/5)}(7/(2+x))

extreme points

f (x) = x^{(\frac{2}{5})} (\frac{7}{2 + x})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

Solution steps

f^{'} (x) = \frac{7 ( - 3 x + 4 )}{5 x ^{\frac{3}{5}} ( x + 2 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

x^{\frac{2}{5}} \frac{7}{2 + x} : 0

Minimum (0, 0)

Plug

x = \frac{4}{3}

into

x^{\frac{2}{5}} \frac{7}{2 + x} : \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10}

Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

Q (x, y) = 5 x + 3 x y

f (x, y) = x^{2} + y^{2} - 10 x + 26

f (x) = x^{3} - 9 x^{2} + 10

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

f (x) = 5 - 2 x^{2}, - 4 \leq x \leq 2

What is the extreme f(x)=x^{(2/5)}(7/(2+x)) ?
The extreme f(x)=x^{(2/5)}(7/(2+x)) is Minimum(0,0),Maximum(4/3 ,(7*4^{2/5}*3^{3/5})/(10))