yEXAMPLE 6Discuss the curve y = 3x* – 48x with respect to concavity, points of inflection, and local maxima and minima. Use this information tosketch the curve.SOLUTIONIf f(x) = 3x4 - 48x³, then144x2 = 12x2(x – 12)288х 3D 36x(х — 8).f'(x)12x3f"(x) = 36x2812To 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 criticalnumbers:f"(0)f"(12)Video ExampleSince f'(12)and f"(12) > 0, f(12) =is a local minimum. Since f"(0)the Second Derivative Test gives no information%3Dabout 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 localmaximum or minimum at 0. [In fact, the expression for f'(x) shows that f decreases to the left of 12 and increase to theof 12.]Since f"(x) = 0 when x = 0 or x =, we divide the real number line into intervals with these numbers as endpoints and complete the followingchart.Intervalf"(x)36x(х — 8) Concavity(-0, 0)upwarddownward+upwardThe point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also,-12288is aninflection point since the curve changes from concave downward to concave upward there.Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in the figure.

If fx=3x4-48x3, thenf'x=12x3-144x2=12x2x-12f''x=36x2-288x=36xx-8 To find the critical numbers we set…

EXAMPLE 6 Discuss the curve y = 3x* - 48x³ with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If f(x) = 3x* – 48x², then f'(x) = 12x3 - 144x² = 12x²(x – 12) f"(x) = 36x2 – 288x = 36x(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 " at these critical numbers: f"(0 : f"(12) Since f'(12) = and f"(12) > 0, (12) = O is a local minimum. Since f"(0) =O, the Second Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for 0

EXAMPLE 6 Discuss the curve y = 3x* - 48x³ with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If f(x) = 3x* – 48x², then f'(x) = 12x3 - 144x² = 12x²(x – 12) f"(x) = 36x2 – 288x = 36x(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 " at these critical numbers: f"(0 : f"(12) Since f'(12) = and f"(12) > 0, (12) = O is a local minimum. Since f"(0) =O, the Second Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for 0

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 94E

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