4092: Use the second derivative test to find the relative extrema

Question

Use the second derivative test to find the relative extrema for f(x) = x⁴ - 8x² - 4.

• Minimum at (2, -20); maximum at (0, -4)
• Minimum at (-2, -20); maximum at (2, -20)
• Minimum at (2, -20) and (-2, -20); maximum at (0, -4)
• Minimum at (0, -4); maximum at (2, -20) and (-2, -12)

 

 

Answer 1

f'(x) = 4x³ - 16x

When f' = 0 we have critical numbers -2, 0, 2.

 

The 2nd derivative test requires plugging the critical numbers into the 2nd derivative.

 

f''(x) = 12x² - 16

 

f''(-2) = 32 > 0 => there is a relative min at -2.

Same for f''(2).

 

f''(0) = -16 < 0 => relative max at 0

 

f(0) = -4, so rel max is (0,-4).

f(-2) = -20 so rel min at (-2,-20)

Ditto for f(2).