What is the extreme f(x)=3x^4+4x^3-12x^2+7 ?

The extreme f(x)=3x^4+4x^3-12x^2+7 is Minimum(-2,-25),Maximum(0,7),Minimum(1,2)

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extreme f(x)=3x^4+4x^3-12x^2+7

extreme points

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 7

Minimum (- 2, - 25), Maximum (0, 7), Minimum (1, 2)

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} + 7 : - 25

Minimum (- 2, - 25)

Plug

x = 0

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 7 : 7

Maximum (0, 7)

Plug

x = 1

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 7 : 2

Minimum (1, 2)

Minimum (- 2, - 25), Maximum (0, 7), Minimum (1, 2)

f (x, y) = 2 x y - 2 x^{2} - y^{3} + 3

f (x, y) = x y e^{(- x - y)}

x^{2} - 6 x + 5

\frac{x}{x + 3}

\frac{1 - x ^{2}}{2 x + 1}

What is the extreme f(x)=3x^4+4x^3-12x^2+7 ?
The extreme f(x)=3x^4+4x^3-12x^2+7 is Minimum(-2,-25),Maximum(0,7),Minimum(1,2)