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

The extreme f(x)=12x^2-3x^4 is Maximum(-sqrt(2),12),Minimum(0,0),Maximum(sqrt(2),12)

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

extreme points

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

Maximum (- 2, 12), Minimum (0, 0), Maximum (2, 12)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

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

Maximum (- 2, 12)

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 2

into

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

Maximum (2, 12)

Maximum (- 2, 12), Minimum (0, 0), Maximum (2, 12)

(- 10, 6), (2, - 4)

\frac{1}{x - 3} + 4

2 x^{2} - 3

1 - x - x^{2}

(8, 0), (10, - 1)

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