What is the extreme f(x)=8+3x^2-x^3 ?

The extreme f(x)=8+3x^2-x^3 is Minimum(0,8),Maximum(2,12)

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extreme f(x)=8+3x^2-x^3

extreme points

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

Minimum (0, 8), Maximum (2, 12)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

8 + 3 x^{2} - x^{3} : 8

Minimum (0, 8)

Plug

x = 2

into

8 + 3 x^{2} - x^{3} : 12

Maximum (2, 12)

Minimum (0, 8), Maximum (2, 12)

f (x) = 4 x^{2} - 16 x + 4, 0 \leq x \leq 3

f (x, y) = 2 x^{2} + 4 x + 2 y^{2} + 5 y

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

- x^{3} + 5 x^{2} + 8 x + 9

- 2 x^{2} + 4 x - 8

What is the extreme f(x)=8+3x^2-x^3 ?
The extreme f(x)=8+3x^2-x^3 is Minimum(0,8),Maximum(2,12)