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Math Help - differentiation

  1. #1
    Member Awsom Guy's Avatar
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    differentiation

    Is this differentiation right.
    question:
    (2x-5)^3

    My Answer:
    6(2x-5)^2

    If I do this how do I find the gradient. If I sub x=0 i don't get what I want.
    Please help.
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  2. #2
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    Quote Originally Posted by Awsom Guy View Post
    Is this differentiation right.
    question:
    (2x-5)^3

    My Answer:
    6(2x-5)^2

    If I do this how do I find the gradient. If I sub x=0 i don't get what I want.
    Please help.
    The differentiation is correct.
    That is the curve gradient for any x.
    If you sub in x=0, you get the slope of the tangent to the curve at x=0,
    but not the slope of the tangent that passes through the origin if that's what you were looking for.

    What were you trying to calculate ?
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  3. #3
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    Quote Originally Posted by Awsom Guy View Post
    Is this differentiation right.
    question:
    (2x-5)^3

    My Answer:
    6(2x-5)^2

    If I do this how do I find the gradient. If I sub x=0 i don't get what I want.
    Please help.
    I assume you have used the chain rule to arrive at this answer.

    If so, then yes, your answer is correct.

    When you sub x = 0 into the derivative, you should find that the gradient at that point is 150.
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  4. #4
    Member Awsom Guy's Avatar
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    My question says: Find the point where the curve y=(2x-5)^3 cuts the x-axis and find the slope of the tangent at this point.
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  5. #5
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    Quote Originally Posted by Awsom Guy View Post
    My question says: Find the point where the curve y=(2x-5)^3 cuts the x-axis and find the slope of the tangent at this point.
    Well then do as you are instructed.

    The find where the curve cuts the x axis, let y = 0.

    So 0 = (2x - 5)^3

    2x - 5 = 0

    x = \frac{5}{2}.


    So you are not evaluating the derivative at x = 0. You are evaluating the derivative at x = \frac{5}{2}.
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  6. #6
    Member Awsom Guy's Avatar
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    do i then place it in the y=mx+b formula to calculate the slope at the tagent at the point.
    Last edited by Awsom Guy; February 9th 2010 at 01:13 AM. Reason: bad spelling :)
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  7. #7
    Member Awsom Guy's Avatar
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    just so everyone knows the answer is m=0
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  8. #8
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    no need,
    the derivative is the slope or gradient of the tangent.
    That's zero when y=0 also in this case.
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  9. #9
    Member Awsom Guy's Avatar
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    But the derivative isn't 0 it is 6(2x-5)^2
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  10. #10
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    It is when x=\frac{5}{2} or y=0 !
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  11. #11
    Member Awsom Guy's Avatar
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    Thank you so much for helping me.
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  12. #12
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    here's the graph,
    the tangent is the x axis itself at x=2.5.
    Attached Thumbnails Attached Thumbnails differentiation-f-x-f-x-0.jpg  
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  13. #13
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    thanks
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