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

The extreme y=x^{2/5}(x+3) is Maximum(-6/7 ,(15(-6/7)^{2/5})/7),Minimum(0,0)

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

extreme points

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

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}, Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{7 x + 6}{5 x ^{\frac{3}{5}}}

Find intervals:

Increasing

: - \infty < x < - \frac{6}{7},

Decreasing

: - \frac{6}{7} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{6}{7}

into

x^{\frac{2}{5}} (x + 3) : \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}

Plug

x = 0

into

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

Minimum (0, 0)

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}, Minimum (0, 0)

f (x) = x^{2} + x + 3

x + 2 cos (x)

f (x) = - 2 x + 3 ln (4 x)

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

f (x, y) = - x^{2} + 5 y^{2} - 7 x + 2 y + 5

What is the extreme y=x^{2/5}(x+3) ?
The extreme y=x^{2/5}(x+3) is Maximum(-6/7 ,(15(-6/7)^{2/5})/7),Minimum(0,0)