Differential Calculus

CHAPTER 1: Complete Solutions

1. .
2. .
3. .
4. .
5. .
6. 2x+h.
7. .
8. (a) , (b) .
9. (a) f(0)=1, (b) f(0.142857)=0.857143, (c) .
10. f(F(x))=x, F(f(x))=|x|.

Exercise Set 2, p.16

1. .
2. All reals except , ...
3. .
4. |x|<2. (You can also write this as -2<x<2.)
5. All reals except
6. .
7. .
8. .
9. .
10. 
11. |x|>1. (That is, either x>1 or x<-1.)
12. . (You can also write this as , or .)
13. The natural domain of f is .
14. .
15. . (You can also write this as |x|<1, or -1<x<1.)

16. 

17. 

18. 

19. 

20. 

21. 

22. 

23. 

24. 

Exercise Set 3, p.26

1. Because -2<0.
2. Because -1/2 is negative.
3. Correction: implies .
4. Correction: A>B>0 implies 1/A<1/B.
5. Correction: A<B implies -A>-B.
6. Correction: If and B>0, then A<B.
7. This statement is correct. There is nothing wrong!
8. . (Note: To complete our argument we need , which is guaranteed by .)
9. It can be bigger than 4, but not bigger than 6.
10. g is unbounded: it can be greater than (resp. less than) any given number.
11. From x>1 we see that both x and x-1 are positive. Hence we can square both sides of the ineqaulity x>x-1 to arrive at . (Alternatively, since both x and x-1 are positive, .)
12. From we see that . Since (certainly this implies the positivity of x), we have , or . Now . So the last inequality can be rewritten as . We can multiply both sides of this inequality by because guarantees that is positive.
13. Since both x and are , we can apply the AG-inequality to get . Since , we have . So .
14. Yes. Under no further condition.
15. Since and , we have , or , which gives . Taking reciprocals, we get . (The last step is legitimate because both and |x| are positive.)
16. |v|<c.

Chapter Exercises, p. 28

1. .
2. .
3. .
4. .
5. From we see that x must be positive: x>0. So we can rewrite it as 3>6x, which gives . Thus the solution is .
6. .
7. .
8. . In other words, either or .
9. . That is, .
10. . That is, x can be any real number.
11. . (Note: For general p, is defined only for z>0.)
12. . Or .

13. 

14. 

15. 

16. 

17. 

18. 

19. 

20. for all x. (Note that is always .)

21. 

22. for all x.
23. From we have . So gives . Now . Thus . On the other hand, from we have . So we can multiply throughout by to arrive at .
24. 2, 2.25, 2.370370, 2.44141, 2.48832, 2.52163, 2.54650, 2.56578, 2.58117, 2.59374.
25. From we have and . Thus we may apply the AG-inequality to get . Multiplying both sides by , we obtain . Using and we will get the required inequality.