Question

Please help me find the inflection point

EXAMPLE 6 Discuss the curve y = 2x4 – 32x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If f(x) = 2x4 – 32x3, then f'(x) 8x3 - 96x2 = 8x2(x – 12) f"(x) = 24x2 192x = 24x(x – 8). To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x = 12 the Second Derivative Test we evaluate f" at these critical numbers: To use f"(0) = 0 f"(12) = 1152 Since f'(12) = 0 and f"(12) > 0, f(12) = 0 x is a local minimum. Since f"(0) = 0 the Second Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for 0<x< 12, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. [In fact, the expression for f'(x) shows that f decreases to the left of 12 and increase to the right of 12.] 1 1 Since f'(x) = 0 when x = 0 or x = 8 we divide the real number line into intervals with these numbers as endpoints and complete the following chart. Interval f"(x) = 24x(x - 8) Concavity (-0,0) + upward 0,8 downward ) 00) 8 + upward The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also, -8192 is an inflection point since the curve changes from concave downward to concave upward there.

Answer 1

please like

Y_322 7 f (x2 = 8x3 -96 x ². (a) - 24x² - 192x 242 Ca-8) - O 8 - a ac Hence point of inflection o are