What is the extreme f(x)=2sqrt(x)-4x,x>0 ?

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

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extreme f(x)=2sqrt(x)-4x,x>0

extreme points

f (x) = 2 x - 4 x, x > 0

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

Solution steps

Find the critical points:

x = \frac{1}{16}

Domain of

2 x - 4 x, x > 0 : x > 0

Find intervals:

Increasing

: 0 < x < \frac{1}{16},

Decreasing

: \frac{1}{16} < x < \infty

Plug the extreme point

x = \frac{1}{16}

into

2 x - 4 x \Rightarrow y = \frac{1}{4}

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

f (x) = e^{x^{2} - 4}, (- 2, 2)

f (x, y) = \frac{y}{x} + x^{2} y + 6 x

y = - \frac{3}{2 x + 2}

4 sin (x) + 5 cos (x)

f (x) = - 3 x^{2} + 24 x - 5

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