What is the extreme f(x)=2sqrt(x)-8x,x>0 ?

The extreme f(x)=2sqrt(x)-8x,x>0 is Maximum(1/64 , 1/8)

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extreme f(x)=2sqrt(x)-8x,x>0

extreme points

f (x) = 2 x - 8 x, x > 0

Maximum (\frac{1}{64}, \frac{1}{8})

Solution steps

Find the critical points:

x = \frac{1}{64}

Domain of

2 x - 8 x, x > 0 : x > 0

Find intervals:

Increasing

: 0 < x < \frac{1}{64},

Decreasing

: \frac{1}{64} < x < \infty

Plug the extreme point

x = \frac{1}{64}

into

2 x - 8 x \Rightarrow y = \frac{1}{8}

Maximum (\frac{1}{64}, \frac{1}{8})

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

y = \frac{1}{x ^{2} - 9}

\frac{9}{x - 4}

f (x) = \frac{x ^{2} - 4 x - 21}{3 x - 21}

y = \frac{x ^{2} - 16}{9 - x ^{2}}

What is the extreme f(x)=2sqrt(x)-8x,x>0 ?
The extreme f(x)=2sqrt(x)-8x,x>0 is Maximum(1/64 , 1/8)