What is the extreme f(x)=4sqrt(x)-6x,x>0 ?

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

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extreme f(x)=4sqrt(x)-6x,x>0

extreme points

f (x) = 4 x - 6 x, x > 0

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

Solution steps

Find the critical points:

x = \frac{1}{9}

Domain of

4 x - 6 x, x > 0 : x > 0

Find intervals:

Increasing

: 0 < x < \frac{1}{9},

Decreasing

: \frac{1}{9} < x < \infty

Plug the extreme point

x = \frac{1}{9}

into

4 x - 6 x \Rightarrow y = \frac{2}{3}

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

f (x) = x (x + 11) (x - 6)

tan (2 x - 5)

f (x) = (x - 2)^{3}

f (x) = - 4 x^{2} + 6 x + 2

f (x) = - x + 3

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