Find all relative extrema. Use the Second-Derivative Test. No credit for other methods. 2) f (x) = 3x^4 - 4x^3 - 12x^2 + 24

Answer 1

The relative extrema are found where f '(x) = 0

So take the first derivative and set it = 0

f '(x) = 12 x^3 - 12 x^2 - 24 x Set this = 0

12 x^3 - 12 x^2 - 24 x = 0 12x [ x^2 - x - 2 ] = 0 12x (x - 2)(x + 1) = 0

x = 0 , x = 2 , x = -1 are the critical values.

For the 2nd derivative test, we need to plug these critical values into the 2nd derivative to determine if the function is concave up at that critical number or concave down. If the function is concave UP, you have found a relative minimum. If the function is concave DOWN, you have found a relative maximum.

f ''(x) = 36 x^2 - 24x - 24 = 12 ( 3x^2 - 2x - 2)

f ''(0) = 12(-2) = negative so concave down so MAXIMUM at x = 0

f ''(2) = 12 (6) = positive so concave up so MINIMUM at x = 2

f ''(-1) = 12 (3) = positive so concave up so MINIMUM at x = -1