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Chapter 4 | Applications of Derivatives
Theorem 4.9: First Derivative Test Suppose that f is a continuous function over an interval I containing a critical point c . If f is differentiable over I , except possibly at point c , then f ( c ) satisfies one of the following descriptions: i. If f ′ changes sign from positive when x < c to negative when x > c , then f ( c ) is a local maximum of f . ii. If f ′ changes sign from negative when x < c to positive when x > c , then f ( c ) is a local minimum of f . iii. If f ′ has the same sign for x < c and x > c , then f ( c ) is neither a local maximum nor a local minimum of f .
We can summarize the first derivative test as a strategy for locating local extrema.
Problem-Solving Strategy: Using the First Derivative Test Consider a function f that is continuous over an interval I . 1. Find all critical points of f and divide the interval I into smaller intervals using the critical points as endpoints. 2. Analyze the sign of f ′ in each of the subintervals. If f ′ is continuous over a given subinterval (which is typically the case), then the sign of f ′ in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point x in that subinterval and by evaluating the sign of f ′ at that test point. Use the sign analysis to determine whether f is increasing or decreasing over that interval. 3. Use First Derivative Test and the results of step 2 to determine whether f has a local maximum, a local minimum, or neither at each of the critical points.
Now let’s look at how to use this strategy to locate all local extrema for particular functions.
Example 4.17 Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for f ( x ) = x 3 −3 x 2 −9 x −1. Use a graphing utility to confirm your results. Solution Step 1. The derivative is f ′( x ) =3 x 2 −6 x −9. To find the critical points, we need to find where f ′( x ) =0. Factoring the polynomial, we conclude that the critical points must satisfy 3( x 2 −2 x −3) =3( x −3)( x +1) =0. Therefore, the critical points are x =3, −1. Now divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 3) and (3, ∞). Step 2. Since f ′ is a continuous function, to determine the sign of f ′( x ) over each subinterval, it suffices to choose a point over each of the intervals (−∞, −1), (−1, 3) and (3, ∞) and determine the sign of f ′ at each
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