Find (by hand) all critical numbers and use the First Derivative Test to classify each as the location of a local maximum, local minimum or neither. y = x^4/3 - 8x^1/3 - 6 A) x = 0 (neither) B) x = -2 (local maxima) C) x = -2 (local minima) D) x = 2 (local minima), x = 0 (neither)

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Quizplus · Accepted Answer

To find critical numbers, we need to find the derivative of the function: y' = (1/3) x^(-2/3) - 8(1/3) x^(-2/3) Setting this equal to zero, we get: 0 = (1/3) x^(-2/3) - 8(1/3) x^(-2/3) Simplifying, we get: 0 = (1/3) x^(-2/3)(1 - 8) 0 = (1/3) x^(-2/3)(-7) We can ignore the constant factor (-7/3) since it will not affect the critical points. So, we have: 0 = x^(-2/3) This means x must be non-zero. Therefore, the only critical points are x = 0 and x = 2. Now, we need to use the First Derivative Test to classify each critical point. For x = 0, we have y' < 0 to the left of x = 0 and y' > 0 to the right of x = 0. Therefore, x = 0 is a local minimum. For x = 2, we have y' < 0 to the left of x = 2 and y' > 0 to the right of x = 2. Therefore, x = 2 is a local minimum. Thus, the best choice is D, as it correctly identifies both local minimums and omits the extraneous critical point at x = -2.

Find (By Hand) All Critical Numbers and Use the First