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

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

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

extreme points

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

Maximum (\frac{1}{4}, \frac{1}{2})

Solution steps

Find the critical points:

x = \frac{1}{4}

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < \infty

Plug the extreme point

x = \frac{1}{4}

into

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

Maximum (\frac{1}{4}, \frac{1}{2})

f (x) = 3 tan (2 x - 8 π) + 3

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

r

6 x^{3}

3 x + 4 y = 24

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