Calculus-Sec-2-2-Solution
2.2 Continuity by SGLee - HSKim -SWSun
http://youtu.be/azrkT1RP4-c, http://youtu.be/hj8d-j_DGf4
1. Given continuous functions 
and 
with 
and 
, find 
.
http://matrix.skku.ac.kr/cal-lab/cal-2-2-1.html
Since 
and 
are continuous functions.
2. Given continuous functions 
and 
with 
and 
, find 
.
Since 
is continuous, 
.
3. Show that the function 
is discontinuous at 
.
http://matrix.skku.ac.kr/cal-lab/cal-2-2-3.html
We may draw 
with the following Sage command. It shows the function diverges
(+infinity) at 
. This shows f(x) is discontinuous at 
.
If we use the following Sage command, we may get limit(abs(ln((x-2)^2))) at 
directly.
It shows 
is discontinuous at 
.
4-7. Determine the points of discontinuity of 
. At which of these numbers is 
continuous from the right, from the left or neither? Sketch the graph of 
.
4. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-4.html
We may draw 
with the following Sage command. It shows the function is discontinuous at 
.
5. 
We see that 
exists for all 
except 
. Notice that the right and left limits are different 
.
6. 
We see that 
exists for all 
except 
and 
. Notice that the right and left limits are different at 
and 
.
7. 
We see that 
exists for all 
except 
and 
. Notice that the right and left limits are different at 
and 
.
8-10. For what values of the constant 
is the function 
continuous on 
?
8. 
Thus, for 
to be continuous on 
10. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-10.html
Suppose that function 
is continuous on 
. Find C such that 
Find C such that 
. 
is continuous at 
(But 
is not continuos at 
). The solution is 
.
11-13. Show that the following functions 
has removable discontinuity at 
. Also find a function 
that agrees with 
for 
and is continuous on ℝ.
11. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-11.html
does not exist. Hence 
has removable discontinuity at 
. Then we can change the expression
to remove the discontinuity.
Let 
. It agrees with 
for 
and is continuous on ℝ.
12. 
for 
.
The discontinuity is removable and 
agrees with 
for 
and is continuous on ℝ.
13. 
for 
.
The discontinuity is removable and 
agrees with 
for 
and is continuous on ℝ.
14. Let 
. Is 
removable discontinuity?
Since 
,
is not a removable discontinuity.
15. If 
, show that there is a number 
such that 
.
is continuous on the interval 
, 
and 
.
Since 
, there is a number 
in 
such that 
by the Intermediate Value Theorem.
16. Prove using Intermediate Value Theorem that there is a positive number 
such that 
.
Let 
. We know that 
is continuous on the interval 
, 
and 
. Since 
,
there is a number 
in 
such that 
by the Intermediate Value Theorem.
17-22. Prove that there is a root of the given equation in the specified interval by using the Intermediate Value Theorem.
17. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-17.html
Let 
. Then 
.
Since 
is a real-valued continuous function on the interval 
, there is a root 
such that 
by the intermediate value theorem.
18. 
, 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-18.html
Let 
. Then 
.
Since 
is a real-valued continuous function on the interval 
, there is a root 
such that 
by the intermediate value theorem.
19. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-19.html
Let 
. Then 
.
Since 
is a real-valued continuous function on the interval 
, there is a root 
such that 
by the intermediate value theorem.
21. 
is continuous on the interval 
, 
and 
.
Since 
, there is a number 
in 
such that 
by the Intermediate Value Theorem.
Thus, there is root of the equation 
in the interval 
.
22. 
is continuous on the interval 
, 
and 
.
Since –6.696<0<e, there is a number 
in 
such that 
by the Intermediate Value Theorem. Thus, there is root of the equation 
in the interval 
.
23-26. Show that each of the following equations has at least one real root.
23. 
http://matrix.skku.ac.kr/cal-lab/cal-2-2-23.html
http://math2.skku.ac.kr/home/pub/37
We may draw both 
and 
in one graph to find intersections. It shows the function has two real roots in 
and 
.
The following Sage commands give the value of the intersection in each interval, 
or 
.
24. 
Let 
. Then 
and 
. So by the Intermediate Value Theorem, there is a number 
in (-1,0) such that 
.
This implies that 
.
25. 
Let 
. Then 
and 
. So by the Intermediate Value Theorem, there is a number 
in (1,2) such that 
.
This implies that 
26. 
Let 
. Then 
and 
, and 
is continuous in 
So by the Intermediate Value Theorem, there is a number 
in 
such that 
. This implies that 
.
27-28. Find the values of 
for which 
is continuous.
27. 
BWOC. Suppose 
is continuous at 
in ℝ, then 
.
But if 
is rational, then 
but 
or 
can be 1. if 
is irrational, then 
but 
or 
can be 0. It occurs a contradiction. So such value 
in ℝ exist.
28. 
is continuous only at 
since 
. But there is no nonzero 
in ℝ. such that 
.
9. 
20. 
, 
