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Math Help - Two Continuity problems

  1. #1
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    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



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    Re: Two Continuity problems

    I hardly can wait to see your efforts, steps and everything, not just the final answer.
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    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~?
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    Re: Two Continuity problems

    Quote Originally Posted by TKHunny View Post
    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

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    Re: Two Continuity problems

    Quote Originally Posted by Hossam View Post
    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~.
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  6. #6
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    Re: Two Continuity problems

    Quote Originally Posted by Plato View Post
    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 !
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  7. #7
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    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.
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  8. #8
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    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 ??
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    Re: Two Continuity problems

    Quote Originally Posted by Plato View Post
    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 !!
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  10. #10
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    Re: Two Continuity problems

    Quote Originally Posted by Hossam View Post
    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.
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  11. #11
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    Re: Two Continuity problems

    Quote Originally Posted by Plato View Post
    You did not answer the question in reply #7.
    answered in reply #9
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  12. #12
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    Re: Two Continuity problems

    This is meant to be honest not mean.
    You are simply are not ready to tackle this problem.
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  13. #13
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    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 ) ??
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  14. #14
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    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.
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  15. #15
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    Re: Two Continuity problems

    Quote Originally Posted by HallsofIvy View Post
    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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