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

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

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

extreme points

f (x) = x^{3} - (\frac{3}{2}) x^{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}{2} x^{2} : 0

Maximum (0, 0)

Plug

x = 1

into

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

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

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

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

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

f (x) = x^{2}, - 1 \leq x \leq 2

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

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

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