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Math Help - Calculas question

  1. #1
    Member Awsom Guy's Avatar
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    Calculas question

    I hope this is the right area to post this question.

    Find the equation of the tangent to the curve y=2/x+1 at the point where x=3.

    I had a go can somebody please check it thanks,

    y=2/x+1
    y=2/4
    =1/2 (3,1/2)

    2(x+1)^-1
    -2(x+1)^-2
    -2/(x+1)^2
    2/(3+1)(3+1)
    2/8
    =1/4
    therefore in y=mx+b form:
    y=1/4x+b
    1/2=1/4(3)+b
    b=2/3
    y=1/4x+2/3


    I have done that and I think I have done it correctly but the answer I have is
    x+8y-7=0
    How can I get this answer or is this wrong.
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  2. #2
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    Quote Originally Posted by Awsom Guy View Post
    I hope this is the right area to post this question.

    Find the equation of the tangent to the curve y=2/x+1 at the point where x=3.

    I had a go can somebody please check it thanks,

    y=2/x+1 I assume you mean 2/(x + 1)
    y=2/4
    =1/2 (3,1/2)

    2(x+1)^-1
    -2(x+1)^-2
    -2/(x+1)^2
    2/(3+1)(3+1) Correct up to here. But remember that 4x4 = 16
    ...
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  3. #3
    Member Awsom Guy's Avatar
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    I still dont get it.

    okay I will re-do that last bit.
    2/16
    1/8
    y=mx+b
    y=1/8x+b
    1/2=1/8(3)+b
    2/3=b

    therefore:
    y=1/8x+1/1/3

    but I still don't get the right answer of x+8y-7=0
    Thanks
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  4. #4
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    Quote Originally Posted by Awsom Guy View Post
    okay I will re-do that last bit.
    2/16
    1/8
    y=mx+b
    y=1/8x+b
    1/2=1/8(3)+b Correct up to here.

    1/2 = 3/8 + b

    b = 1/2 - 3/8

    b = 4/8 - 3/8

    b = 1/8.


    So y = (1/8)x + 1/8

    8y = x + 1.


    I disagree with the answer you have been given. Did you copy it down correctly?
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  5. #5
    Member Awsom Guy's Avatar
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    Thanks

    Quote Originally Posted by Prove It View Post
    I disagree with the answer you have been given. Did you copy it down correctly?
    [/COLOR][/COLOR]
    Thanks for that.
    The answer was in a maths booklet I have the exact question with me. I am not sure why the answer is wrong,I'll make sure I recheck everything.
    Thanks
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  6. #6
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    Actually, I figured out what's wrong.

    You have \frac{dy}{dx} = -\frac{2}{(x + 1)^2}.

    So at x = 3 you have

    \frac{dy}{dx} = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8}, not \frac{1}{8}.


    The rest should follow.
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