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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

x^{3} - \frac{3 x ^{2}}{2} : 0

Maximum (0, 0)

Plug

x = 1

into

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

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

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

f (x) = 2 x - 4 cos (x)

f (x) = 16 x y - x^{3} - 8 y^{2}

f (x) = x + \frac{9}{x} + 3

f (x) = - x^{4} + 8 x^{2} + 10

f (x) = (x^{\frac{5}{2} - 1} \cdot e^{- \frac{x}{2}})

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