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Chapter 4 | Applications of Derivatives
Step 5. Calculate the first derivative:
2 (1)
2 −2 x ( x −1) 2 .
f ′( x ) = ( x −1)(2 x )− x
= x
( x −1) 2
We have f ′( x ) =0 when x 2 −2 x = x ( x −2) =0. Therefore, x =0 and x =2 are critical points. Since f is undefined at x =1, we need to divide the interval (−∞, ∞) into the smaller intervals (−∞, 0), (0, 1), (1, 2), and (2, ∞), and choose a test point from each interval to evaluate the sign of f ′( x ) in each of these smaller intervals. For example, let x =−1, x = 1 2 , x = 3 2 , and x =3 be the test points as shown in the following table. Interval Test Point Sign of f ′ ( x ) = x 2 −2 x ( x −1 ) 2 = x ( x −2 ) ( x −1 ) 2 Conclusion
(−∞, 0)
(−)(−)/ + = +
x =−1
f is increasing.
(0, 1)
(+)(−)/ + = −
x =1/2
f is decreasing.
(1, 2)
(+)(−)/ + = −
x =3/2
f is decreasing.
(2, ∞)
(+)(+)/ + = +
x =3
f is increasing.
From this table, we see that f has a local maximum at x =0 and a local minimum at x =2. The value of f at the local maximum is f (0) =0 and the value of f at the local minimum is f (2) =4. Therefore, (0, 0) and
(2, 4) are important points on the graph. Step 6. Calculate the second derivative:
( x −1) 2 (2 x −2)− ⎛
⎝ x 2 −2 x ⎞ ⎠ ⎛
⎝ 2( x −1) ⎞ ⎠
f ″( x ) =
( x −1) 4
⎡ ⎣ ( x −1)(2 x −2)−2 ⎛
⎝ x 2 −2 x ⎞ ⎠ ⎤ ⎦
( x −1)
=
( x −1) 4
( x −1)(2 x −2)−2 ⎛
⎝ x 2 −2 x ⎞ ⎠
=
( x −1) 3
2 x 2 −4 x +2− ⎛
⎝ 2 x 2 −4 x ⎞ ⎠
=
( x −1) 3
= 2
( x −1) 3 .
We see that f ″( x ) is never zero or undefined for x in the domain of f . Since f is undefined at x =1, to check concavity we just divide the interval (−∞, ∞) into the two smaller intervals (−∞, 1) and (1, ∞), and choose a test point from each interval to evaluate the sign of f ″( x ) in each of these intervals. The values x =0
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