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

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

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extreme f(x)=2sqrt(x)-8x

extreme points

f (x) = 2 x - 8 x

Maximum (0, 0), Minimum (\frac{1}{64}, \frac{1}{8})

Solution steps

f^{'} (x) = \frac{1}{x} - 8

Find intervals:

Decreasing

: 0 \leq x < 64,

Increasing

: 64 < x < \infty

Plug

x = 0

into

2 x - 8 x : 0

Maximum (0, 0)

Plug

x = \frac{1}{64}

into

2 x - 8 x : \frac{1}{8}

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

Maximum (0, 0), Minimum (\frac{1}{64}, \frac{1}{8})

\frac{1}{x ^{3}}

f (x) = \frac{x ^{7} - 5}{3} + 3

x + 2 y = - 2

1 - x - - x^{2} + 5 x - 4

\frac{( \frac{x}{x + 2} )}{( \frac{x}{x + 2} ) + 2}

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