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

The extreme f(x)=x^3-(3x^2)/2+3 is Maximum(0,3),Minimum(1, 5/2)

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

extreme points

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

Maximum (0, 3), Minimum (1, \frac{5}{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} + 3 : 3

Maximum (0, 3)

Plug

x = 1

into

x^{3} - \frac{3 x ^{2}}{2} + 3 : \frac{5}{2}

Minimum (1, \frac{5}{2})

Maximum (0, 3), Minimum (1, \frac{5}{2})

f (x) = 200 + 8 x^{3} + x^{4}

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 24

f (x) = x^{3} - 6 x^{2} + 9 x + 15

f (x, y) = 2 x + 2 x y + y

x^{2} y - x^{2} - 2 y^{2}

What is the extreme f(x)=x^3-(3x^2)/2+3 ?
The extreme f(x)=x^3-(3x^2)/2+3 is Maximum(0,3),Minimum(1, 5/2)