Determine where the function f(x)=(x2−1)2/3 is increasing and where it is decreasing.
Solution f is continuous for all numbers x, and f′(x)=23(x2−1)−1/3(2x)=4x3(x2−1)1/3
The Increasing/Decreasing Function Test states that f is increasing on intervals where f′(x)>0 and decreasing on intervals where f′(x)<0. We solve these inequalities by using the numbers −1, 0, and 1 to form four intervals. Then we determine the sign of f′ in each interval, as shown in Table 2. We conclude that f is increasing on the intervals (−1,0) and (1,∞) and that f is decreasing on the intervals (−∞,−1) and (0,1).
Interval | Sign of 4x | Sign of (x2−1)1/3 | Sign of f′(x)=4x3(x2−1)1/3 | Conclusion |
---|---|---|---|---|
(−∞,−1) | Negative (−) | Positive (+) | Negative (−) | f is decreasing on (−∞,−1) |
(−1,0) | Negative (−) | Negative (−) | Positive (+) | f is increasing on (−1,0) |
(0,1) | Positive (+) | Negative (−) | Negative (−) | f is decreasing on (0,1) |
(1,∞) | Positive (+) | Positive (+) | Positive (+) | f is increasing on (1,∞) |