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Chapter 4 | Applications of Derivatives
2
⎛ ⎝ 1− x 2
⎞ ⎠
⎛ ⎝ 2
⎛ ⎝ 1− x 2
⎞ ⎠ (−2 x )
⎞ ⎠
(2)−2 x
f ″( x ) =
4
⎛ ⎝ 1− x 2
⎞ ⎠
⎡ ⎣ 2
⎤ ⎦
⎛ ⎝ 1− x 2
⎞ ⎠
⎛ ⎝ 1− x 2
⎞ ⎠ +8 x 2
=
4
⎛ ⎝ 1− x 2 ⎞ ⎠ +8 x 2 ⎞ ⎠
⎛ ⎝ 1− x 2
2
=
3
⎛ ⎝ 1− x 2
⎞ ⎠
2 +2
= 6 x
3 .
⎛ ⎝ 1− x 2
⎞ ⎠
To determine the intervals where f is concave up and where f is concave down, we first need to find all points x where f ″( x ) =0 or f ″( x ) is undefined. Since the numerator 6 x 2 +2≠0 for any x , f ″( x ) is never zero. Furthermore, f ″ is not undefined for any x in the domain of f . However, as discussed earlier, x =±1 are not in the domain of f . Therefore, to determine the concavity of f , we divide the interval (−∞, ∞) into the three smaller intervals (−∞, −1), (−1, −1), and (1, ∞), and choose a test point in each of these intervals to evaluate the sign of f ″( x ). in each of these intervals. The values x =−2, x =0, and x =2 are possible test points as shown in the following table. Interval Test Point Sign of f ″ ( x ) = 6 x 2 +2 ⎛ ⎝ 1− x 2 ⎞ ⎠ 3 Conclusion
(−∞, −1)
x =−2
+/− =−
f is concave down.
(−1, −1)
x =0
+/+ = +
f is concave up.
(1, ∞)
x =2
+/− =−
f is concave down.
Combining all this information, we arrive at the graph of f shown below. Note that, although f changes concavity at x =−1 and x =1, there are no inflection points at either of these places because f is not continuous at x =−1 or x =1.
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