Find the local maximum value(s) of $y = x^{4} - 6 x^{2}$

If

f (x) = x^{4} - 6 x^{2}

, then

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

f'' (x) = 12 x^{2} - 12 = 12 (x^{2} - 1)

To find the critical numbers we set

f' (x) = 0

and obtain

x = 0, x = \sqrt{3}

and

x = - \sqrt{3}

. Now we use the Second Derivative Test to evaluate

f''

at these critical numbers:

f'' (0) = - 12 < 0

,

f'' (\sqrt{3}) = 24 > 0

, and

f'' (- \sqrt{3}) = 24 > 0

Since f' (0)=0 and

f'' (0) < 0, f (0) = 0

is a local maximum. Since

f' (\sqrt{3}) = 0

and

f'' (\sqrt{3}) > 0, f (\sqrt{3}) = - 9

is a local minimum. Similarly, since

f' (- \sqrt{3}) = 0

and

f'' (- \sqrt{3}) > 0, f (- \sqrt{3}) = - 9

is a local minimum.

If

f (x) = x^{4} - 6 x^{2}