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Chapter 4 | Applications of Derivatives
4.16 Use the first derivative test to locate all local extrema for f ( x ) =− x 3 + 3 2
x 2 +18 x .
Example 4.18 Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for f ( x ) =5 x 1/3 − x 5/3 . Use a graphing utility to confirm your results.
Solution Step 1. The derivative is
⎛ ⎝ 1− x 4/3 ⎞ ⎠
5
2/3 3 =
5−5 x 4/3 3 x 2/3
− 5 x
x −2/3 − 5 3
x 2/3 = 5
f ′( x ) = 5 3
=
.
3 x 2/3
3 x 2/3
The derivative f ′( x ) =0 when 1− x 4/3 =0. Therefore, f ′( x ) =0 at x =±1. The derivative f ′( x ) is undefined at x =0. Therefore, we have three critical points: x =0, x =1, and x =−1. Consequently, divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞). Step 2: Since f ′ is continuous over each subinterval, it suffices to choose a test point x in each of the intervals from step 1 and determine the sign of f ′ at each of these points. The points x =−2, x = − 1 2 , x = 1 2 , and x =2 are test points for these intervals. Interval Test Point Sign of f ′ ( x ) = 5 ⎛ ⎝ 1− x 4/3 ⎞ ⎠ 3 x 2/3 at Test Point Conclusion
(+)(−)
(−∞, −1)
x =−2
f is decreasing.
+ =−
(+)(+)
(−1, 0)
f is increasing.
x = − 1 2
+ = +
(+)(+)
(0, 1)
f is increasing.
x = 1 2
+ = +
(+)(−)
(1, ∞)
x =2
f is decreasing.
+ =−
Step 3: Since f is decreasing over the interval (−∞, −1) and increasing over the interval (−1, 0), f has a local minimum at x =−1. Since f is increasing over the interval (−1, 0) and the interval (0, 1), f does not have a local extremum at x =0. Since f is increasing over the interval (0, 1) and decreasing over the interval (1, ∞), f has a local maximum at x =1. The analytical results agree with the following graph.
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