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Chapter 4 | Applications of Derivatives
x −1) 2/3 =∞, the function continues to grow without bound as x →∞ and x →−∞.
Step 3: Since lim x →±∞ (
Step 4: The function has no vertical asymptotes. Step 5: To determine where f is increasing or decreasing, calculate f ′. We find f ′( x ) = 2 3 ( x −1) −1/3 = 2 3( x −1) 1/3 .
This function is not zero anywhere, but it is undefined when x =1. Therefore, the only critical point is x =1. Divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞), and choose test points in each of these intervals to determine the sign of f ′( x ) in each of these smaller intervals. Let x =0 and x =2 be the test points as shown in the following table. Interval Test Point Sign of f ′ ( x ) = 2 3 ( x −1 ) 1/3 Conclusion
(−∞, 1)
x =0
+/− =−
f is decreasing.
(1, ∞)
x =2
+/+ = +
f is increasing.
We conclude that f has a local minimum at x =1. Evaluating f at x =1, we find that the value of f at the local minimum is zero. Note that f ′(1) is undefined, so to determine the behavior of the function at this critical point, we need to examine lim x →1 f ′( x ). Looking at the one-sided limits, we have lim x →1 + 2 3( x −1) 1/3 =∞and lim x →1 − 2 3( x −1) 1/3 =−∞. Therefore, f has a cusp at x =1. Step 6: To determine concavity, we calculate the second derivative of f : f ″( x ) = − 2 9 ( x −1) −4/3 = −2 9( x −1) 4/3 . We find that f ″( x ) is defined for all x , but is undefined when x =1. Therefore, divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞), and choose test points to evaluate the sign of f ″( x ) in each of these intervals. As we did earlier, let x =0 and x =2 be test points as shown in the following table. Interval Test Point Sign of f ″ ( x ) = −2 9 ( x −1 ) 4/3 Conclusion
(−∞, 1)
x =0
−/+ =−
f is concave down.
(1, ∞)
x =2
−/+ =−
f is concave down.
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