Calculus-Sec-4-1-Solution
4.1 Extreme Values of a Function by SGLee - HSKim-SWSun
1- 4. Determine if the following statement is true or false. Explain your answer.
1. If 
is a continuous function, then 
has a maximum at only one point in 
.
False. Consider 
that has more than one maximum.
2. One can apply the Mean Value Theorem to 
on 
.
False, since 
is not differentiable on the open interval 
.
3. If a continuous function 
has a extreme value on 
, then 
has a absolute maximum or minimum value on 
.
False. Draw the graph of 
on 
.
4. For a continuous function 
, 
has only one zero provided 
is strictly decreasing.
True. We may prove it By the Way Of Contradiction (BWOC).
5- 8. Find all critical numbers of given functions.
5. 
.
http://matrix.skku.ac.kr/cal-lab/cal-4-1-5.html
6. 
Therefore critical numbers of 
are 
.
To find critical numbers, we differentiate 
.
Then 
.
Therefore, 
and 
(since 
).
So, critical numbers of 
are 
or 
.
9-13. Find all local maxima and minima if there exist.
9. 
.
http://matrix.skku.ac.kr/cal-lab/cal-4-1-9.html
No local minimum and maximum when 
x=0 is a critical point, but have not local maximum or local minimum.
Thus, when x=pi/2, f(pi/2)=1 is a local maximum.
12. 
if 
, 
, then 
has a local minimum at 
.
if 
, 
, then 
has a local minimum at 
.
if 
, 
, then 
has a local maximum at 
.
13. 
, 
then 
has a local maximum at 
14-17. Find the intervals where the function is increasing or decreasing.
14. 
Decreasing on 
, increasing on 
16. 
Answer : Increasing on 
, decreasing on 
17. Suppose 
and 
for all 
. Using the Mean Value Theorem show 
for all 
.
By the Mean Value Theorem, 
and 
Since 
, we have
.
18-20. Prove the inequality.
18. 
So, 
for 
19. 
, 
Let 
, 
, 
. Then 
, 
,
. 
From the inequality 
and the Mean Value Theorem,
, 
. Therefore 
for all 
.
20. 
and 
So, 
for all 
.
21. Let 
. Find the number of zeros and give a proper interval containing all zeros.
http://matrix.skku.ac.kr/cal-lab/cal-4-1-20.html
7 zeros in 
22. Show that 
for 
.
Then, 
.
23-24. Prove the inequality using the Mean Value Theorem.
23. 
.
Let 
, then by the Mean Value Theorem there exists 
in 
such that
. Since 
for 
, we have
.
24. 
Let 
, than there exists 
in 
such that
. Since,
, we have 
.
25. Prove that the inequality 
for all 
and 
using the Mean Value Theorem.
Let 
, 
, than there exists 
in 
such that
Thus,
.
26. Let 
be a continuous function. Prove that the only functions satisfying 
for all 
are of the form 
, by the Mean Value Theorem.
Applying Mean Value Theorem to 
on 
, there exists 
in 
such that
for all 
. This implies
Now set 
, 
, then 
for all 
.
27. Suppose 
is continuous on 
and differentiable on 
. Prove that there exists 
such that 
.
Let 
, then
.
By Rolle's theorem, here exist 
such that 
, so
.
28. Prove Fermat's Theorem: If 
is differentiable at 
and has an extreme value at 
, then 
.
http://en.wikipedia.org/wiki/Fermat's_theorem_(stationary_points)
Suppose, for the sake of definiteness, that 
has a local maximum at 
. Then, 
if 
is sufficiently close to 
.
This implies that if 
is sufficiently close to 0, with 
being positive or negative, then 
and therefore 
.
We can divide both sides of this inequality by a positive number. Thus, if 
and 
is sufficiently small, we have
.
Taking the right-hand limit of both sides of this inequality, we get
Since 
exists, we have
and so we have shown that 
.
If 
, then the direction of the inequality (1) is reversed. We have
.
So, taking the left-hand limit, we have
We have shown that 
and also that 
. Since both of these inequalities must be true, the only possibility is that 
.
29. Prove Rolle's Theorem. (http://en.wikipedia.org/wiki/Rolle's_theorem)
Since 
, we can use the above result due to Farmat’s Theorem to conclude this.
Suppose that the maximum is obtained at an interior point 
of 
. We shall examine the above right- and left-hand limits separately.
For a real 
such that 
is in 
, the value 
is smaller or equal to 
because 
attains its maximum at 
. Therefore, for every 
, 
, hence
, where the limit exists by assumption (it may be minus infinity).
Similarly, for every 
, the inequality is reversed because the denominator is now negative and we get
,
hence
, where the limit might be plus infinity.
Finally, because the above right- and left-hand limits agree, the derivative of 
at 
must be zero.
30. Prove the Increasing and Decreasing Test.
(a) Let 
and 
be any two numbers in the interval with 
. According to the definition of an increasing function we have to show that 
.
Applying the Mean Value theorem to 
on the interval 
, there exists 
such that 
.
By assumptions, 
and 
, we get 
. In particular, 
. Since 
and 
were arbitrary, we conclude that 
whenever 
.
31. Prove the cases (ii), (iii) of the First Derivative Test. (http://en.wikipedia.org/wiki/First_derivative_test)
(ii) Let us choose 
sufficiently near to 
such that 
. By the Mean Value Theorem, there exist 
with 
and 
such that
and 
.
This implies 
for 
and 
for 
. In particular 
is decreasing on 
and increasing on 
. Hence 
has a local minimum at 
.
(iii) Let us choose 
sufficiently near 
such that 
. By the Mean Value Theorem, there exist 
with 
and 
such that
and 
.
Since 
for 
and 
for 
. Hence 
has neither local maximum nor local minimum at 
.
32. Prove the case (b) of the Second Derivative Test. (http://en.wikipedia.org/wiki/Second_derivative_test)
(b) Suppose we have 
. Then
Thus, for 
sufficiently small we get 
which means that 
if 
so that 
is decreasing to the left of 
, and that 
if 
so that 
is increasing to the right of 
.
Now, by the first derivative test we know that 
has a local maximum at 
.
7. 
.
8. 
.
10. 
.
Local minimum 
, but no local maximum.
11. 
.
15. 
Increasing on 
, decreasing on 
