# Math Help - Two Continuity problems

1. ## Two Continuity problems

I want the solution for the 2 questions please , i want the full solution ( with steps ) not just final answers .... thx in advance

2. ## Re: Two Continuity problems

I hardly can wait to see your efforts, steps and everything, not just the final answer.

3. ## Re: Two Continuity problems

The way it works here is that you show us where or what you do not understand. Show some effort.

HINT: If $f$ is an odd function, can you show $f(0)=0~?$

4. ## Re: Two Continuity problems

Originally Posted by TKHunny
I hardly can wait to see your efforts, steps and everything, not just the final answer.

Do u mean u want to see if i tried to solve them ?? , i can't solve the first problem at all but i tried in the second one , tell me if my solution is correct please

5. ## Re: Two Continuity problems

Originally Posted by Hossam
Do u mean u want to see if i tried to solve them ?? , i can't solve the first problem at all but i tried in the second one , tell me if my solution is correct please
Yes that is what mean. However I cannot read any of that graphic.
Can you post in LaTeX?
At least post the proof of
If $f$ is an odd function, show that $f(0)=0~.$

6. ## Re: Two Continuity problems

Originally Posted by Plato
Yes that is what mean. However I cannot read any of that graphic.
Can you post in LaTeX?
At least post the proof of
If $f$ is an odd function, show that $f(0)=0~.$

i dont know how to prove it because i dont know what f(x) is equal to in the problem ?? .. say f(x) = x , then f(0) = 0 and f(-x) = 0 ... therefore f(x)+f(-x)=0+0=0 ... thats all i know about odd functions .. i dont understand how to solve the quest !

7. ## Re: Two Continuity problems

Can you follow this?
Suppose that $f(0)\ne 0$ then either $f(0)<0\text{ or }f(0)>0$.

If $f(0)<0$ then $-f(0)>0$ but $f(0)=f(-0)=-f(0)>0$.

That is a contradiction.

The same is true if $f(0)>0$ so $f(0)=0$.

8. ## Re: Two Continuity problems

in the second question , it says the function is discontinuous at c=2 , now f(2) = 4B-A = ??? , is it equal to 0 or 6 ??

9. ## Re: Two Continuity problems

Originally Posted by Plato
Can you follow this?
Suppose that $f(0)\ne 0$ then either $f(0)<0\text{ or }f(0)>0$.

If $f(0)<0$ then $-f(0)>0$ but $f(0)=f(-0)=-f(0)>0$.

That is a contradiction.

The same is true if $f(0)>0$ so $f(0)=0$.
i understand , u mean if F(x) = x-2 then f(0) <0 , and if F(x)=x+2 then f(0) > 0 , but i cant get how to use this to solve my question !!

10. ## Re: Two Continuity problems

Originally Posted by Hossam
in the second question , it says the function is discontinuous at c=2 , now f(2) = 4B-A = ??? , is it equal to 0 or 6 ??
You did not answer the question in reply #7.

11. ## Re: Two Continuity problems

Originally Posted by Plato
You did not answer the question in reply #7.

12. ## Re: Two Continuity problems

This is meant to be honest not mean.
You are simply are not ready to tackle this problem.

13. ## Re: Two Continuity problems

okay what about the other problem ? , it says fn is discon. at c=2 , in this case does f(2) = 0 , because i know it doesnt equal ( lim 3x at x=2 ) ??

14. ## Re: Two Continuity problems

You cannot find specific values of A and B in (2) and the problem does not ask you to. Saying that the function is continuous at x= 1 tells you that the two one sided limits are equal: A- B= 3. Saying that the function is not continuous at x= 2 tells you that the two one side limits there are not equal. From the left, the limit is 3(2)= 6. From the right, it is 4B- A and we can only say that $4B- A\ne 6$. From, A- B= 3, we get B= A- 3 and we can put that into the inequality: A- B= A- (A- 3)[/tex] which reduces to just 3. Certainly 3 is NOT equal to 6 so the only "condition" we need for A and B is that A- B= 3. What the value of the function is at x= 2 depends upon what A and B individulally are. For example, if we take A= 3, then B= 0 satisfies A- B= 3. In that case, the value of f(2) is 4B- A= -3. But we could also take A= 4, B= 1. That also satifies A- B= 3. Now the value of f(2) is 4B- A= 4-4= 0.

15. ## Re: Two Continuity problems

Originally Posted by HallsofIvy
You cannot find specific values of A and B in (2) and the problem does not ask you to. Saying that the function is continuous at x= 1 tells you that the two one sided limits are equal: A- B= 3. Saying that the function is not continuous at x= 2 tells you that the two one side limits there are not equal. From the left, the limit is 3(2)= 6. From the right, it is 4B- A and we can only say that $4B- A\ne 6$. From, A- B= 3, we get B= A- 3 and we can put that into the inequality: A- B= A- (A- 3)[/tex] which reduces to just 3. Certainly 3 is NOT equal to 6 so the only "condition" we need for A and B is that A- B= 3. What the value of the function is at x= 2 depends upon what A and B individulally are. For example, if we take A= 3, then B= 0 satisfies A- B= 3. In that case, the value of f(2) is 4B- A= -3. But we could also take A= 4, B= 1. That also satifies A- B= 3. Now the value of f(2) is 4B- A= 4-4= 0.
ahaaa got it ,, thxxx for ur support

but what about da first problem it's getting me maaad !!

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