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

The extreme f(x)=7x^{2/5}+2x^{7/5} is Maximum(-1,5),Minimum(0,0)

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

extreme points

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

Maximum (- 1, 5), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{14 + 14 x}{5 x ^{\frac{3}{5}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 5)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 1, 5), Minimum (0, 0)

f (x) = (ln (x))^{2}

f (x) = (\frac{4860}{x}) + 12 x + 709407

\frac{10000}{2.5 cos ( x ) + sin ( x )}

f (x) = (x - 9) e^{x}

f (x) = 3 x, 27 \leq x \leq 64

What is the extreme f(x)=7x^{2/5}+2x^{7/5} ?
The extreme f(x)=7x^{2/5}+2x^{7/5} is Maximum(-1,5),Minimum(0,0)