What is the extreme f(x)=sqrt(x)((x^2)/5-4) ?

The extreme f(x)=sqrt(x)((x^2)/5-4) is Maximum(0,0),Minimum(2,-(16sqrt(2))/5)

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extreme f(x)=sqrt(x)((x^2)/5-4)

extreme points

f (x) = x (\frac{x ^{2}}{5} - 4)

Maximum (0, 0), Minimum (2, - \frac{16 2}{5})

Solution steps

f^{'} (x) = \frac{( x + 2 ) ( x - 2 )}{2 x}

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x (\frac{x ^{2}}{5} - 4) : 0

Maximum (0, 0)

Plug

x = 2

into

x (\frac{x ^{2}}{5} - 4) : - \frac{16 2}{5}

Minimum (2, - \frac{16 2}{5})

Maximum (0, 0), Minimum (2, - \frac{16 2}{5})

- x^{2} - 6 x - 4

f (x) = \frac{x}{x - 9}, 11 \leq x \leq 13

f (x) = x 1 - x

f (x) = \frac{x + 1}{2 x + x + 1}

f (x, y) = 7 x^{2} + 8 x y + 9 y^{2} + 9 x y^{2} + 5 x^{2} y

What is the extreme f(x)=sqrt(x)((x^2)/5-4) ?
The extreme f(x)=sqrt(x)((x^2)/5-4) is Maximum(0,0),Minimum(2,-(16sqrt(2))/5)