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Math Help - Tangent Line Question

  1. #1
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    Tangent Line Question

    Hi, I have a quick question about finding the equation of a tangent line to the graph of y = (3x+5)e^x at x=0

    so far I have Point (0,5)
    Code:
    (3(x+h) +5)e^(x+h) - (3x+5)e^x
    --------------------------------
                   h
    
    (3x+3h+5)e^x + e^h - (3x+5)e^x
    --------------------------------
                   h
    would I factor for the next step?
    Last edited by drewms64; October 31st 2006 at 04:51 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by drewms64 View Post
    Hi, I have a quick question about finding the equation of a tangent line to the graph of y = (3x+5)e^x at x=0

    so far I have Point (0,5)
    Code:
    (3(x+h) +5)e^(x+h) - (3x+5)e^x
    --------------------------------
                   h
    
    (3x+3h+5)e^x + e^h - (3x+5)e^x
    --------------------------------
                   h
    would I factor for the next step?
    To start with you have an error in your second line. It should read:

    \frac{(3x+3h+5)e^xe^h - (3x+5)e^x}{h}

    I presume you are calculating the first derivative longhand?

    I would actually rewrite the second line as:

    \frac{(3x+5)e^xe^h - (3x+5)e^x + 3he^xe^h}{h}

    (3x+5)e^x \left ( \frac{e^h-1}{h} \right ) + 3e^xe^h

    When you take your limit as h goes to 0 the second term is easy. What you have left is to figure out what
    \lim_{h \to 0} \frac{e^h-1}{h}

    For the record, the limit is 1. If you need help with it, just let me know.

    -Dan
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  3. #3
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    Ah thanks I missed that error, also are there any other ways to go from my second step onward? I'm just unsure of the way you did it

    Edit: such as maybe
    Code:
    3xe^h + 3he^h +5e^h
    ----------------------
                  h
    although im not sure if that would work
    Last edited by drewms64; October 31st 2006 at 06:14 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by drewms64 View Post
    Ah thanks I missed that error, also are there any other ways to go from my second step onward? I'm just unsure of the way you did it

    Edit: such as maybe
    Code:
    3xe^h + 3he^h +5e^h
    ----------------------
                  h
    although im not sure if that would work
    Factoring the e^x out of the whole expression would also work. It's just a matter of taste.

    Quote Originally Posted by topsquark View Post
    To start with you have an error in your second line. It should read:
    \frac{(3x + 3h + 5)e^xe^h - (3x+5)e^x}{h}
    What I am doing between these lines is simply rearranging the terms so I can factor a (3x+5)e^x from the first two terms.

    \frac{(3x+5)e^xe^h - (3x+5)e^x + 3he^xe^h}{h}

    \frac{ (3x+5)e^x (e^h - 1) + 3he^xe^h}{h} <-- Here's the factorization.

    Now I'm going to split the fraction into two terms and divide by h:
    (3x+5)e^x \left ( \frac{e^h-1}{h} \right ) + 3e^xe^h
    Hope that helps. Again the explicit way you break the problem up doesn't matter. You'll get the same result in the end.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Factoring the e^x out of the whole expression would also work. It's just a matter of taste.
    Yeah that helps showing me how you did it your way. If I was to do it by factoring out the e^x, how would I go about canceling out the H's from the e^h's? This way I can plug in 2 to find the slope to get the formula for the tangent line.
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