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Chapter 4 | Applications of Derivatives
x →±∞. If g ( x ) is a linear function, it is known as an oblique asymptote . 4. Determine whether f has any vertical asymptotes. 5. Calculate f ′. Find all critical points and determine the intervals where f is increasing and where f is decreasing. Determine whether f has any local extrema. 6. Calculate f ″. Determine the intervals where f is concave up and where f is concave down. Use this information to determine whether f has any inflection points. The second derivative can also be used as an alternate means to determine or verify that f has a local extremum at a critical point.
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Example 4.28 Sketching a Graph of a Polynomial
Sketch a graph of f ( x ) = ( x −1) 2 ( x +2).
Solution Step 1. Since f is a polynomial, the domain is the set of all real numbers. Step 2. When x =0, f ( x ) =2. Therefore, the y -intercept is (0, 2). To find the x -intercepts, we need to solve the equation ( x −1) 2 ( x +2) =0, gives us the x -intercepts (1, 0) and (−2, 0) Step 3. We need to evaluate the end behavior of f . As x →∞, ( x −1) 2 →∞ and ( x +2)→∞. Therefore, lim x →∞ f ( x ) =∞. As x →−∞, ( x −1) 2 →∞ and ( x +2)→−∞. Therefore, lim x →−∞ f ( x ) =−∞. To get even more information about the end behavior of f , we can multiply the factors of f . When doing so, we see that f ( x ) = ( x −1) 2 ( x +2) = x 3 −3 x +2. Since the leading term of f is x 3 , we conclude that f behaves like y = x 3 as x →±∞. Step 4. Since f is a polynomial function, it does not have any vertical asymptotes. Step 5. The first derivative of f is f ′( x ) =3 x 2 −3. Therefore, f has two critical points: x =1, −1. Divide the interval (−∞, ∞) into the three smaller intervals: (−∞, −1), (−1, 1), and (1, ∞). Then, choose test points x =−2, x =0, and x =2 from these intervals and evaluate the sign of f ′( x ) at each of these test points, as shown in the following table.
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