What is the extreme f(x)=4sqrt(x)-6x ?

The extreme f(x)=4sqrt(x)-6x is Maximum(0,0),Minimum(1/9 , 2/3)

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extreme f(x)=4sqrt(x)-6x

extreme points

f (x) = 4 x - 6 x

Maximum (0, 0), Minimum (\frac{1}{9}, \frac{2}{3})

Solution steps

f^{'} (x) = \frac{2}{x} - 6

Find intervals:

Decreasing

: 0 \leq x < 9,

Increasing

: 9 < x < \infty

Plug

x = 0

into

4 x - 6 x : 0

Maximum (0, 0)

Plug

x = \frac{1}{9}

into

4 x - 6 x : \frac{2}{3}

Minimum (\frac{1}{9}, \frac{2}{3})

Maximum (0, 0), Minimum (\frac{1}{9}, \frac{2}{3})

x^{3} - 6

f (x) = 4 x

f (x) = 5 x^{2} + 2

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

\frac{tan ( θ )}{θ}

What is the extreme f(x)=4sqrt(x)-6x ?
The extreme f(x)=4sqrt(x)-6x is Maximum(0,0),Minimum(1/9 , 2/3)