What is the extreme f(x)=x^{2/9}(2x+11) ?

The extreme f(x)=x^{2/9}(2x+11) is Maximum(-1,9),Minimum(0,0)

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extreme f(x)=x^{2/9}(2x+11)

extreme points

f (x) = x^{\frac{2}{9}} (2 x + 11)

Maximum (- 1, 9), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{22 ( x + 1 )}{9 x ^{\frac{7}{9}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

x^{\frac{2}{9}} (2 x + 11) : 9

Maximum (- 1, 9)

Plug

x = 0

into

x^{\frac{2}{9}} (2 x + 11) : 0

Minimum (0, 0)

Maximum (- 1, 9), Minimum (0, 0)

\frac{- x ^{6} - 5 x ^{3} + 5 x}{x ^{2} + 2}

f (x) = x^{3} - 3 x^{2} - 9 x + 1, - 2 \leq x \leq 2

f (x) = x ln (8 x), (0, \infty)

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

f (x) = 19 + 2 x - x^{2}, 0 \leq x \leq 5

What is the extreme f(x)=x^{2/9}(2x+11) ?
The extreme f(x)=x^{2/9}(2x+11) is Maximum(-1,9),Minimum(0,0)