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Math Help - Find an equation for the line tangent to the graph

  1. #1
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    Find an equation for the line tangent to the graph

    Find an equation for the line tangent to the graph of f(x)=3x(e^x) at the point (a, f(a)) if a=2.

    I think you start with taking a derivative. Isn't the derivative just f'(x)=3e^x or do you have to do the product rule? After that, I'm not sure what to do.
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  2. #2
    Super Member wingless's Avatar
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    f(x) = 3xe^x

    \frac{df}{dx} = \frac{d}{dx}3xe^x = 3 \frac{d}{dx}xe^x~ .....

    Now use the product rule
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  3. #3
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    Ok, so I took the derivative and got f'(x)=3xe^x + 3e^x.

    I'm just not sure about how to formulate the equation.
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  4. #4
    Super Member wingless's Avatar
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    The slope of the tangent line of the function f(x) at the point x=a is f'(a).

    General formula of a line is y=mx + C. m is the slope (which is also f'(a) for the tangent line) and C is a constant. You can find C by plugging x=a and y=f(a) in the formula.
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    Sorry to keep asking questions, but I worked it out and still didn't get it right.

    So I plugged 2 into the derivative and got 66.50150489, and then I plugged in a and f(a) and found C to be .3334012213, so I thought my equation would be
    y=66.50150489(x+.3334012213), but the math program I'm using is telling me that this is incorrect.
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  6. #6
    Super Member wingless's Avatar
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    Probably because the equation is y = mx + C rather than y = m (x + C) as you used.
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    Even when I used y=66.50150489x + .3334012213, it still says that's incorrect.
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  8. #8
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    Quote Originally Posted by A5HLEY View Post
    Find an equation for the line tangent to the graph of f(x)=3x(e^x) at the point (a, f(a)) if a=2.

    I think you start with taking a derivative. Isn't the derivative just f'(x)=3e^x or do you have to do the product rule? After that, I'm not sure what to do.
    Let's start from the beginning:

    f(x)=3x \cdot e^x~\implies~ f'(x)=3x \cdot e^x + 3e^x = 3e^x(x+1)

    If a = 2 then P(2, 6e^2) and the slope of the tangent is:

    m = f'(2)=3 \cdot e^2 \cdot (2+1)=9e^2

    You know a point and the slope of the tangent. Use the point-slope-formula to calculate the equation of the straight line:

    y - 6e^2 = 9e^2(x-2)~\implies~\boxed{y = 9e^2 \cdot x - 12e^2}

    As you may have noticed: The terms I use are a lot shorter than the decimal numbers you use - and they are exact! The decimal numbers are only approximative values.
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  9. #9
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    Thank you. You're right... your values make it much easier to understand. Thank you again for your help!
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