What is the extreme f(x)=x^2-sqrt(2x)+1/2 ?

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

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extreme f(x)=x^2-sqrt(2x)+1/2

extreme points

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

Maximum (0, \frac{1}{2}), Minimum (\frac{1}{2}, - \frac{1}{4})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}

Domain of

x^{2} - 2 x + \frac{1}{2} : x \geq 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug the extreme point

x = 0

into

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

Maximum (0, \frac{1}{2})

Plug the extreme point

x = \frac{1}{2}

into

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

Minimum (\frac{1}{2}, - \frac{1}{4})

Maximum (0, \frac{1}{2}), Minimum (\frac{1}{2}, - \frac{1}{4})

\frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x

f (x, y) = x^{3} - y^{3} - 3 x y

f (x) = x^{2} + y^{2} + 1

f (x) = 3 x^{2} + 2 x^{3}

f (x) = - 5 x^{2} + 50 x - 80

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