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

The extreme f(x)=2sqrt(x)-4x is Maximum(0,0),Minimum(1/16 , 1/4)

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extreme f(x)=2sqrt(x)-4x

extreme points

f (x) = 2 x - 4 x

Maximum (0, 0), Minimum (\frac{1}{16}, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{1}{x} - 4

Find intervals:

Decreasing

: 0 \leq x < 16,

Increasing

: 16 < x < \infty

Plug

x = 0

into

2 x - 4 x : 0

Maximum (0, 0)

Plug

x = \frac{1}{16}

into

2 x - 4 x : \frac{1}{4}

Minimum (\frac{1}{16}, \frac{1}{4})

Maximum (0, 0), Minimum (\frac{1}{16}, \frac{1}{4})

f (x) = \frac{1}{2} x^{3} - 3 x y + 3 y^{2} - x

\frac{x ^{2}}{( x - 1 ) ^{2}}

f (x) = \frac{3}{2} x^{4} - x^{6}

f (x, y) = - 3 x^{2} + 7 x y - 4 y^{2} + x + y

x^{4} + x^{3} - 6 x^{2}

What is the extreme f(x)=2sqrt(x)-4x ?
The extreme f(x)=2sqrt(x)-4x is Maximum(0,0),Minimum(1/16 , 1/4)