What is the extreme f(x)=x(2x-3)^2 ?

The extreme f(x)=x(2x-3)^2 is Maximum(1/2 ,2),Minimum(3/2 ,0)

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extreme f(x)=x(2x-3)^2

extreme points

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

Maximum (\frac{1}{2}, 2), Minimum (\frac{3}{2}, 0)

Solution steps

f^{'} (x) = 3 (2 x - 3) (2 x - 1)

Find intervals:

Increasing

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

Decreasing

: \frac{1}{2} < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = \frac{1}{2}

into

x (2 x - 3)^{2} : 2

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

Plug

x = \frac{3}{2}

into

x (2 x - 3)^{2} : 0

Minimum (\frac{3}{2}, 0)

Maximum (\frac{1}{2}, 2), Minimum (\frac{3}{2}, 0)

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

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

f (x) = 2 x^{3} - 33 x^{2} + 168 x - 20

y = sin (x)

g (x, y) = e^{y} (y - x^{2})

What is the extreme f(x)=x(2x-3)^2 ?
The extreme f(x)=x(2x-3)^2 is Maximum(1/2 ,2),Minimum(3/2 ,0)