Chapter 4 | Applications of Derivatives
427
Interval
Test Point
Sign of Derivative f ′ ( x ) =3 x 2 −3=3 ( x −1 )( x +1 )
Conclusion
(−∞, −1)
(+)(−)(−) = +
x =−2
f is increasing.
(−1, 1)
(+)(−)(+) = −
x =0
f is decreasing.
(1, ∞)
(+)(+)(+) = +
x =2
f is increasing.
From the table, we see that f has a local maximum at x =−1 and a local minimum at x =1. Evaluating f ( x ) at those two points, we find that the local maximum value is f (−1) =4 and the local minimum value is f (1) =0. Step 6. The second derivative of f is f ″( x ) =6 x . The second derivative is zero at x =0. Therefore, to determine the concavity of f , divide the interval (−∞, ∞) into the smaller intervals (−∞, 0) and (0, ∞), and choose test points x =−1 and x =1 to determine the concavity of f on each of these smaller intervals as shown in the following table.
Sign of f ″ ( x ) =6 x
Interval
Test Point
Conclusion
(−∞, 0)
x =−1
−
f is concave down.
(0, ∞)
x =1
+
f is concave up.
We note that the information in the preceding table confirms the fact, found in step 5, that f has a local maximumat x =−1 and a local minimum at x =1. In addition, the information found in step 5 —namely, f has a local maximum at x =−1 and a local minimum at x =1, and f ′( x ) =0 at those points—combined with the fact that f ″ changes sign only at x =0 confirms the results found in step 6 on the concavity of f . Combining this information, we arrive at the graph of f ( x ) = ( x −1) 2 ( x +2) shown in the following graph.
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