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

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

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

extreme

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 1

into

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

Minimum (1, - 5)

Plug

x = 3

into

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

Maximum (3, 27)

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

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