What is the extreme f(x)=-3x^4+8x^3+18x^2 ?

The extreme f(x)=-3x^4+8x^3+18x^2 is Maximum(-1,7),Minimum(0,0),Maximum(3,135)

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extreme f(x)=-3x^4+8x^3+18x^2

extreme

Maximum (- 1, 7), Minimum (0, 0), Maximum (3, 135)

Solution steps

f^{'} (x) = - 12 x^{3} + 24 x^{2} + 36 x

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 1

into

- 3 x^{4} + 8 x^{3} + 18 x^{2} : 7

Maximum (- 1, 7)

Plug

x = 0

into

- 3 x^{4} + 8 x^{3} + 18 x^{2} : 0

Minimum (0, 0)

Plug

x = 3

into

- 3 x^{4} + 8 x^{3} + 18 x^{2} : 135

Maximum (3, 135)

Maximum (- 1, 7), Minimum (0, 0), Maximum (3, 135)

What is the extreme f(x)=-3x^4+8x^3+18x^2 ?
The extreme f(x)=-3x^4+8x^3+18x^2 is Maximum(-1,7),Minimum(0,0),Maximum(3,135)