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Thomas Calculus 13th [Solutions]

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Chapter 4 Practice Exercises 319<br />

12. The absolute maximum is | 1| 1 and the absolute minimum is |0| 0. This is not inconsistent with the<br />

Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains<br />

its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an<br />

interval which is half open and half closed, such as [ 1,1), so there is nothing to contradict.<br />

13. (a) There appear to be local minima at x 1.75<br />

and 1.8. Points of inflection are indicated at<br />

approximately x 0 and x 1.<br />

(b)<br />

(c)<br />

7 5 4<br />

f ( x) x 3x 5x<br />

indicates a local maximum at<br />

2 2 2 3<br />

15 x x ( x 3)( x 5). The pattern<br />

x 3 5 and local minima at x 3.<br />

y<br />

|<br />

3<br />

| | |<br />

0<br />

3<br />

5 3<br />

14. (a) The graph does not indicate any local<br />

extremum. Points of inflection are indicated<br />

at approximately x 3 and x 1.<br />

4<br />

(b) f ( x) x 7 2x<br />

4 5 10<br />

3<br />

x<br />

local maximum at<br />

(c)<br />

3 3 7<br />

x ( x 2)( x 5). The pattern<br />

x 7 5 and a local minimum at<br />

x<br />

3 2.<br />

f<br />

)( |<br />

0<br />

7<br />

5<br />

| indicates a<br />

3 2<br />

Copyright<br />

2014 Pearson Education, Inc.

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