Chapter 4 | Applications of Derivatives
429
x 2 1− x 2
x 2 1− x 2
lim x →1 +
=−∞and lim x →1 −
=∞.
In addition, by looking at each one-sided limit as x →−1, we find that lim x →−1 + x 2 1− x 2 =∞and lim x →−1 − x 2 1− x 2
=−∞.
Step 5. Calculate the first derivative:
⎛ ⎝ 1− x 2
⎞ ⎠ (2 x )− x 2 (−2 x ) ⎛ ⎝ 1− x 2 ⎞ ⎠ 2
x
f ′( x ) =
= 2
2 .
⎛ ⎝ 1− x 2
⎞ ⎠
Critical points occur at points x where f ′( x ) =0 or f ′( x ) is undefined. We see that f ′( x ) =0 when x =0. The derivative f ′ is not undefined at any point in the domain of f . However, x =±1 are not in the domain of f . Therefore, to determine where f is increasing and where f is decreasing, divide the interval (−∞, ∞) into four smaller intervals: (−∞, −1), (−1, 0), (0, 1), and (1, ∞), and choose a test point in each interval to determine the sign of f ′( x ) in each of these intervals. The values x =−2, x = − 1 2 , x = 1 2 , and x =2 are good choices for test points as shown in the following table. Interval Test Point Sign of f ′ ( x ) = 2 x ⎛ ⎝ 1− x 2 ⎞ ⎠ 2 Conclusion
(−∞, −1)
x =−2
−/+ =−
f is decreasing.
(−1, 0)
x =−1/2
−/+ =−
f is decreasing.
(0, 1)
x =1/2
+/+ = +
f is increasing.
(1, ∞)
x =2
+/+ = +
f is increasing.
From this analysis, we conclude that f has a local minimum at x =0 but no local maximum. Step 6. Calculate the second derivative:
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