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

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

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

extreme

Maximum (- \frac{2}{7}, 3 (- \frac{2}{7})^{\frac{2}{3}}), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{5 ( 7 x + 2 )}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < - \frac{2}{7},

Decreasing

: - \frac{2}{7} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{2}{7}

into

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

Maximum (- \frac{2}{7}, 3 (- \frac{2}{7})^{\frac{2}{3}})

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- \frac{2}{7}, 3 (- \frac{2}{7})^{\frac{2}{3}}), Minimum (0, 0)

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