What is the extreme f(x)=x^{2/5}(3x+7) ?

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

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extreme f(x)=x^{2/5}(3x+7)

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{2}{3}

into

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

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

Plug

x = 0

into

x^{\frac{2}{5}} (3 x + 7) : 0

Minimum (0, 0)

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

x^{3} - 2 x^{2} - 15 x + 10 [- 2.1]

f (x) = x + \frac{9}{x} + 2

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 30

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

+ 2.5 t^{2} + 6 t

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