What is the extreme f(x)=3x^4-4x^3-36x^2+5 ?

The extreme f(x)=3x^4-4x^3-36x^2+5 is Minimum(-2,-59),Maximum(0,5),Minimum(3,-184)

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extreme f(x)=3x^4-4x^3-36x^2+5

extreme

Minimum (- 2, - 59), Maximum (0, 5), Minimum (3, - 184)

Solution steps

f^{'} (x) = 12 x^{3} - 12 x^{2} - 72 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

3 x^{4} - 4 x^{3} - 36 x^{2} + 5 : - 59

Minimum (- 2, - 59)

Plug

x = 0

into

3 x^{4} - 4 x^{3} - 36 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 3

into

3 x^{4} - 4 x^{3} - 36 x^{2} + 5 : - 184

Minimum (3, - 184)

Minimum (- 2, - 59), Maximum (0, 5), Minimum (3, - 184)

What is the extreme f(x)=3x^4-4x^3-36x^2+5 ?
The extreme f(x)=3x^4-4x^3-36x^2+5 is Minimum(-2,-59),Maximum(0,5),Minimum(3,-184)