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Chapter 13: Problem 6

Explain how the Second Derivative Test is used.

Short Answer

Expert verified
Answer: The critical points of the function f(x) = x^3 - 3x^2 are x = 0 (local maximum) and x = 2 (local minimum).

Step by step solution

01

Understand the Objective of the Second Derivative Test

The Second Derivative Test is used to classify the critical points of a function. A critical point occurs when the first derivative (f'(x)) is either equal to zero or is undefined. The critical points can be local maxima, local minima, or inflection points. The Second Derivative Test checks the concavity of the function at the critical points to determine their classification. If f''(x) > 0, the point is a local minimum; if f''(x) < 0, the point is a local maximum; if f''(x) = 0 or is undefined, the test is inconclusive.
02

Find the First and Second Derivatives of the Function

Determine the first derivative, f'(x), by differentiating the given function, f(x), with respect to x. Then, determine the second derivative, f''(x), by differentiating f'(x) with respect to x. Simplify the expressions if possible.
03

Locate the Critical Points

Find the critical points by setting the first derivative, f'(x), equal to zero or locating the points where f'(x) is undefined. Solve for x, and keep the resulting values as the critical points.
04

Apply the Second Derivative Test

Evaluate the second derivative, f''(x), for each critical point found in Step 3. If f''(x) > 0, the point is a local minimum; if f''(x) < 0, the point is a local maximum; if f''(x) = 0 or is undefined, the test is inconclusive.
05

Example: Using the Second Derivative Test

Suppose we have the function f(x) = x^3 - 3x^2. 1. Calculate the first and second derivatives of the function: f'(x) = 3x^2 - 6x and f''(x) = 6x - 6 2. Find the critical points by setting f'(x) = 0: 3x^2 - 6x = 0 3x(x-2) = 0 So, we have two critical points: x = 0 and x = 2 3. Apply the Second Derivative Test: At x = 0: f''(0) = (6(0) - 6) = -6 Since f''(0) < 0, x = 0 is a local maximum. At x = 2: f''(2) = (6(2) - 6) = 6 Since f''(2) > 0, x = 2 is a local minimum. The critical points of the function f(x) = x^3 - 3x^2 are x = 0 (local maximum) and x = 2 (local minimum), determined using the Second Derivative Test.

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