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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.
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