Calculus-Sec-2-2-Solution

2.2   Continuity                                        by SGLee - HSKim -SWSun

1. Given continuous functions  and  with  and , find .

Since  and  are continuous functions.

2. Given continuous functions  and  with  and , find .

Since  is continuous, .

3. Show that the function is discontinuous at .

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.

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.

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.

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.

Let . Then .

Since  is a real-valued continuous function on the interval , there is a root  such that  by the intermediate value theorem.

18.

Let . Then .

Since  is a real-valued continuous function on the interval , there is a root  such that  by the intermediate value theorem.

19.

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.

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 .

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