What is the extreme f(x)=3x^4-16x^3+18x^2 ?

The extreme f(x)=3x^4-16x^3+18x^2 is Minimum(0,0),Maximum(1,5),Minimum(3,-27)

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extreme f(x)=3x^4-16x^3+18x^2

extreme points

f (x) = 3 x^{4} - 16 x^{3} + 18 x^{2}

Minimum (0, 0), Maximum (1, 5), Minimum (3, - 27)

Solution steps

f^{'} (x) = 12 x^{3} - 48 x^{2} + 36 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

3 x^{4} - 16 x^{3} + 18 x^{2} : 0

Minimum (0, 0)

Plug

x = 1

into

3 x^{4} - 16 x^{3} + 18 x^{2} : 5

Maximum (1, 5)

Plug

x = 3

into

3 x^{4} - 16 x^{3} + 18 x^{2} : - 27

Minimum (3, - 27)

Minimum (0, 0), Maximum (1, 5), Minimum (3, - 27)

\frac{1}{x + 3} + 2

f (x) = - x^{99}

2 x^{3} - 1

x^{2} - x - 2

12 y^{4} + 24 y - 10

What is the extreme f(x)=3x^4-16x^3+18x^2 ?
The extreme f(x)=3x^4-16x^3+18x^2 is Minimum(0,0),Maximum(1,5),Minimum(3,-27)