utilizes the second derivative test to classify

Question

Utilizes the second derivative test to classify the critical points of the function,

                                              h ( x ) = 3x⁵ - 5x³ + 3

Solution

The second derivative is,

                           h′′ ( x ) + 60x³ - 30x

The three critical points (x = -1, x = 0, & x = 1) of this function are all critical points where the first derivative is zero therefore we know that at least we have a possibility that the Second Derivative Test will work. For each of these the value of the second derivative is,

                h′′ ( -1) = -30                            h′′ (0) = 0              h′′ (1) = 30

The second derivative at x = -1 is -ve the second derivative at x = 1 is +ve and therefore we have a relative minimum here by the Second Derivative Test

In the case of x = 0 the second derivative is zero and therefore we can't utilizes the Second Derivative Test to classify this critical point.