What is the extreme f(x)=12x^4-4x^2+6 ?

The extreme f(x)=12x^4-4x^2+6 is Minimum(-1/(sqrt(6)), 17/3),Maximum(0,6),Minimum(1/(sqrt(6)), 17/3)

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extreme f(x)=12x^4-4x^2+6

extreme

Minimum (- \frac{1}{6}, \frac{17}{3}), Maximum (0, 6), Minimum (\frac{1}{6}, \frac{17}{3})

Solution steps

f^{'} (x) = 48 x^{3} - 8 x

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{6},

Increasing

: - \frac{1}{6} < x < 0,

Decreasing

: 0 < x < \frac{1}{6},

Increasing

: \frac{1}{6} < x < \infty

Plug

x = - \frac{1}{6}

into

12 x^{4} - 4 x^{2} + 6 : \frac{17}{3}

Minimum (- \frac{1}{6}, \frac{17}{3})

Plug

x = 0

into

12 x^{4} - 4 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = \frac{1}{6}

into

12 x^{4} - 4 x^{2} + 6 : \frac{17}{3}

Minimum (\frac{1}{6}, \frac{17}{3})

Minimum (- \frac{1}{6}, \frac{17}{3}), Maximum (0, 6), Minimum (\frac{1}{6}, \frac{17}{3})

What is the extreme f(x)=12x^4-4x^2+6 ?
The extreme f(x)=12x^4-4x^2+6 is Minimum(-1/(sqrt(6)), 17/3),Maximum(0,6),Minimum(1/(sqrt(6)), 17/3)