What is the extreme f(x)=3x^4+4x^3-12x^2+5 ?

The extreme f(x)=3x^4+4x^3-12x^2+5 is Minimum(-2,-27),Maximum(0,5),Minimum(1,0)

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extreme f(x)=3x^4+4x^3-12x^2+5

extreme points

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 5

Minimum (- 2, - 27), Maximum (0, 5), Minimum (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 5 : - 27

Minimum (- 2, - 27)

Plug

x = 0

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 1

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 5 : 0

Minimum (1, 0)

Minimum (- 2, - 27), Maximum (0, 5), Minimum (1, 0)

arccos (\frac{2 x + 1}{x - 3})

f (x) = 7 - 6 x^{2}

f (x) = (x - 1)^{3}

f (x) = \frac{x - 4}{3 x - x ^{2}}

y = 6

What is the extreme f(x)=3x^4+4x^3-12x^2+5 ?
The extreme f(x)=3x^4+4x^3-12x^2+5 is Minimum(-2,-27),Maximum(0,5),Minimum(1,0)