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Math Help - Find Equation of a Line Tangent

  1. #1
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    Find Equation of a Line Tangent

    I need help finding the line tangent to the graph of the following:

    f(x)=4x^3-2x^2+5x-3

    at the point x=6

    Please help me with this; it's confusing!
    So far I took the derivative of the equation and got:
    12x^2-4x+5

    Where should I go from here?

    Any help is appreciated!
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  2. #2
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    ok, so f'(x)=12x^2-4x+5
    Now you want to plug in x=6 to find the slope of the tangent line:
    12(6)^2-4(6)+5
    =432-24+5=413

    Now we're going to find the y coordinate when x=6
    f(x)=4x^3-2x^2+5x-3
    y=4(6)^3-2(6)^2+5(6)-3=819

    so your point is (6,819)

    Now use that to find the equation for the tangent line:
    y-819=413(x-6)
    y-819=413x-2487
    y=413x-1659

    I'm pretty sure that's right. If anybody sees something I missed feel free to comment.
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  3. #3
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    Thank You!

    Awesome! Thank you Climberboy123! That was what I was thinking about doing (substitution 6 into the derived equation), but I wasn't sure.

    Thanks again!
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  4. #4
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    One last question

    I have one last problem:

    Find the derivative:
    y=sqrt(x)*(x+34)

    I know that:
    (dy/dx)x=1
    and
    (dy/dx)34=0
    but I am uncertain about (dy/dx)sqrt[x].

    So would I have (so far):
    y=(dy/dx)sqrt[x]*1
    =(dy/dx)sqrt[x]
    ?

    Where do I go from here?
    Again, help is appreciated!
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  5. #5
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    You took the derivative of part of a product, which you can't do. The only way I can see to solve it is to use the product rule.

    the sqrt(x)=x^(1/2), so the derivative would be (1/2)*x^(-1/2) which =1/(2*sqrt(x))

    If you know how to use the product rule for derivatives, I would use that.
    It's:
    (dy/dx)[f(x)*g(x)]=g'(x)*f(x)+f'(x)*g(x)
    I hope that helps. If that doesn't make sense I can work it out.
    Last edited by Climberboy123; October 2nd 2009 at 03:18 PM.
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  6. #6
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    If you wouldn't mind working it out, Climberboy, I'd appreciate it very much! The only rules I know of so far are the power rule (or is this similar to the product rule?), linear functions, and sums/differences. If there is a way these can be used (in combination) to solve this, please work it out for me. Appreciate the replies!
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  7. #7
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    ok, I just took another look at the problem and you can do it without the product rule.
    So first we want to multiply it all out:

    f(x)=sqrt(x)*(x+34)=x^(1/2)*(x+34)=x^(1/2)*x+x^(1/2)*34

    =x^(3/2)+34x^(1/2)

    So, now we can use the power rule to find f'(x)

    f'(x)=(3/2)x^(1/2)+(34/2)x^(-1/2)= (3x^(1/2))/2+34/(2x^(1/2))

    multiply the first part by [x^(1/2)]/[x^(1/2)] to get a common denominator:

    =[(3x^(1/2))/2]*[x^(1/2)]/[x^(1/2)]+34/(2x^(1/2))

    =(3x)/(2x^(1/2))+34/(2x^(1/2))

    =(3x+34)/(2x^(1/2))

    =(3x+34)/(2*sqrt(x))

    I hope you can follow that. It's kind of hard to read with all the brackets.
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  8. #8
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    I'll have to look that over tomorrow, as my brain doesn't quite work well late at night

    Thanks for the replies, I'm sure it will make sense to me as I look it over!
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