Discuss the curve y = 2x^4 - 32x^3 with respect to concavity, points of inflecti

Question

Discuss the curve y = 2x^4 - 32x^3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. If f(x) = 2x^4 - 32z^3, then f'(x) = 8x^3 - 96x^2 = 8x^2(x - 12) f"(x) = 24x^2 - 192x = 24x(x - 8). To find the critical numbers we set f '(x) = 0 and obtain x = 0 and x = |. To use the Second Derivative Test we evaluate f" at these critical numbers: f"(0) =0 f"(12) = Since f'(12)= and f"(12) > 0, f(12) = is a local minimum. Since f"(0) =, the Second Derivative Test gives no information about the critical number 0. But since f'(x)

Explanation / Answer

1.

x = 0, and x = 12

2.

f''(12) = 24*12*(12 - 8) = 1152

3.

since f'(12) = 0 and f''(12) > 0, f(12) = 2*12^4 - 32*12^3 = -13824 is a local minimum. since f''(0) = 0, the second derivative test gives no information.

4.

since f''(x) = 0, when x = 0 and x = 8, we divide the real number line into intervals.....

5.

(-infinity, 0) + upward

(0, 8) - downward

(8, infinity) + upward

6.

Also (8, -8192) is an inflection point since the curve changes from concave downward.......

Let me know if you have any doubt. And all the bolded letters are your answers.

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