Using the First Derivative Test for Local Extrema Find the critical points of
EXAMPLE 2 Using the First Derivative Test for Local Extrema Find the critical points of
ƒsxd = x 1 >3 sx - 4d = x 4 >3 - 4x 1 >3 .
Identify the intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values.
Solution
The function ƒ is continuous at all x since it is the product of two continuous
functions, x 1 >3 and sx - 4d . The first derivative
is zero at x= 1 and undefined at x= 0. There are no endpoints in the domain, so the crit- ical points x= 0 and x= 1 are the only places where ƒ might have an extreme value. The critical points partition the x-axis into intervals on which ƒ¿ is either positive or negative. The sign pattern of ƒ¿ reveals the behavior of ƒ between and at the critical points. We can display the information in a table like the following:
Intervals
x6 0 06x61
x7 1
Sign of ƒ¿
Behavior of ƒ
decreasing
decreasing
increasing
266
Chapter 4: Applications of Derivatives
Corollary 3 to the Mean Value Theorem tells us that ƒ decreases on s - q , 0d , de-
4 creases on (0, 1), and increases on s1, q d . The First Derivative Test for Local Extrema tells us that ƒ does not have an extreme value at x= 0 ( ƒ¿ does not change sign) and that ƒ
y ⫽x 1/3 (x ⫺ 4)
has a local minimum at x= 1 ( ƒ¿ changes from negative to positive).
2 The value of the local minimum is ƒs1d = 1 1 >3 s1 - 4d = - 3 . This is also an ab-
1 solute minimum because the function’s values fall toward it from the left and rise away
–1 0 1 2 3 4 from it on the right. Figure 4.24 shows this value in relation to the function’s graph.
–1
Note that lim x:0 ƒ¿sxd = - q , so the graph of ƒ has a vertical tangent at the origin.
–2 –3
(1, ⫺3) FIGURE 4.24 The function ƒsxd =
x 1/3 sx - 4d decreases when x6 1 and increases when x7 1 (Example 2).
Chapter 4: Applications of Derivatives
EXERCISES 4.3
18. g sxd = x 4 - 4x 3 + 4x Analyzing ƒ Given ƒⴕ 2
17. ƒsxd = x 4 - 8x 2 + 16
Answer the following questions about the functions whose derivatives 19. Hstd = 3 t 4 -t 6 20. Kstd = 2 15t 3 -t are given in Exercises 1–8: 5 a. What are the critical points of ƒ?
21. g sxd = x 2 8-x 2 22. g sxd = x 2 25 - x b. On what intervals is ƒ increasing or decreasing?
24. x 3 c. At what points, if any, does ƒ assume local maximum and
23. x ƒsxd = 2 -3 xZ2
x- 2 ,
ƒsxd = 3x 2 +1
minimum values?
26. g sxd = x 2 >3 sx + 5d 1. ƒ¿sxd = xsx - 1d
25. ƒsxd = x 1 >3 sx + 8d
28. k sxd = x 2 >3 sx 2 - 4d 3. ƒ¿sxd = sx - 1d 2 sx + 2d
2. ƒ¿sxd = sx - 1dsx + 2d
27. hsxd = x 1 >3 sx 2 - 4d
4. ƒ¿sxd = sx - 1d 2 sx + 2d 2
5. ƒ¿sxd = sx - 1dsx + 2dsx - 3d
Extreme Values on Half-Open Intervals
6. ƒ¿sxd = sx - 7dsx + 1dsx + 5d 7. ƒ¿sxd = x -1 >3 sx + 2d
8. ƒ¿sxd = x -1 >2
In Exercises 29–36:
sx - 3d
a. Identify the function’s local extreme values in the given domain,
Extremes of Given Functions
and say where they are assumed.
In Exercises 9–28: b. Which of the extreme values, if any, are absolute? a. Find the intervals on which the function is increasing and
T c. Support your findings with a graphing calculator or computer decreasing.
grapher.
29. ƒsxd = 2x - x , -q6x…2
b. Then identify the function’s local extreme values, if any, saying
where they are taken on.
30. ƒsxd = sx + 1d 2 , -q6x…0
c. Which, if any, of the extreme values are absolute? 31. g sxd = x 2 - 4x + 4, 1…x6q T d. Support your findings with a graphing calculator or computer
32. g sxd = - x 2 - 6x - 9, -4 … x 6 q grapher.
33. ƒstd = 12t - t 3 , -3 … t 6 q
9. g std = - t 2 - 3t + 3
10. g std = - 3t 2 + 9t + 5
34. ƒstd = t 3 - 3t 2 , -q6t…3
2 3 3 35. hsxd = x 3 - 2x 2 + 4x, 13. 0…x6q ƒsud = 3u - 4u 14. ƒsud = 6u - u 15. ƒsrd = 3r 3 + 16r
11. hsxd = - x 3 + 2x 2 12. hsxd = 2x 3 - 18x
16. hsrd = sr + 7d 3 36. k sxd = x 3 + 3x 2 + 3x + 1, -q6x…0
267 Graphing Calculator or Computer Grapher
4.3 Monotonic Functions and The First Derivative Test
c. ƒ¿sxd 7 0 for x Z 1 ;
In Exercises 37–40:
d. ƒ¿sxd 6 0 for x Z 1 .
a. Find the local extrema of each function on the given interval, and 44. Sketch the graph of a differentiable function y= ƒsxd that has say where they are assumed.
a. a local minimum at (1, 1) and a local maximum at (3, 3); T b. Graph the function and its derivative together. Comment on the
b. a local maximum at (1, 1) and a local minimum at (3, 3); behavior of ƒ in relation to the signs and values of ƒ¿ .
c. local maxima at (1, 1) and (3, 3);
37. x
ƒsxd = x -2 sin , 0 … x … 2p
d. local minima at (1, 1) and (3, 3).
2 2 2 45. Sketch the graph of a continuous function y = g sxd 38. such that ƒsxd = - 2 cos x- cos x , -p … x … p
a. gs
2d = 2, - 0 6 g¿ 6 1 for x 6 2, g¿sxd : 1 as
x:2 + + , -1 6 g¿ 6 0 for x 7 2, and g¿sxd : -1 as 40. ƒsxd = sec 2
39. ƒsxd = csc 2 x- 2 cot x , 06x6p
g¿ 6 0 for x 6 2, g¿sxd : - q as + x:2 , g¿ 7 0 for x 7 2, and g¿sxd : q as x : 2 .
Theory and Examples
46. Sketch the graph of a continuous function y = hsxd such that Show that the functions in Exercises 41 and 42 have local extreme val-
ues at the given values of u , and say which kind of local extreme the 0d = 0, - 2 … hsxd … 2 for all x, h¿sxd : q as x : 0 + ,
a. hs
and h¿sxd : q as x : 0 ;
function has.
u 0d = 0, - 2 … hsxd … 0 for all x, h¿sxd : q as x : 0 + ,
b. hs
2 , 0 … u … 2p, at u = 0 and u = 2p 47. As x moves from left to right through the point c= 2, is the
41. hsud = 3 cos and h¿sxd : - q as x : 0 .
42. 3 hsud = u 5 sin graph of ƒsxd = x - 3x + 2 rising, or is it falling? Give reasons 2 , 0 … u … p, at u = 0 and u = p
for your answer.
43. Sketch the graph of a differentiable function y= ƒsxd through 48. Find the intervals on which the function ƒsxd = ax 2 + bx + c, the point (1, 1) if ƒ¿s1d = 0 and
aZ 0, is increasing and decreasing. Describe the reasoning be- a. ƒ¿sxd 7 0 for x 6 1 and ƒ¿sxd 6 0 for x 7 1 ;
hind your answer.
b. ƒ¿sxd 6 0 for x 6 1 and ƒ¿sxd 7 0 for x 7 1 ;
4.4 Concavity and Curve Sketching