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Math Help - Finding the Equation of Tangent Line

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    Finding the Equation of Tangent Line

    Find an equation of the tangent line to the graph of f at the indicated point
    f(x)=x2+1
    (2, 5)

    This is what I did:
    First step - lim (x+△x)2+1-(x+△x)^2)
    △x->0 .........................△x
    Plugged in 0 to every △x, which turns out to be 1/0, which is undefined

    Second step - (x+△x)(x+△x)+1-(x+△x)(x+△x)
    ....................................△x
    Turns into x2+2△x+△x2+1-x2-2△x-△x2
    ....................................△x
    I got lost here, I think I messed up on the 2nd parenthesis, because I thought I should have turned everything negative since there was a minus sign.
    But then it cancels everything else except the +1

    Can someone help me from here?
    Thanks
    Last edited by Chaim; October 1st 2012 at 03:54 PM.
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    Re: Finding the Equation of Tangent Line

    f(x) = x^2+1

    f(x + \Delta x) = (x+\Delta x)^2 + 1 (you forgot the 1)

    try again ...
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by Chaim View Post
    Find an equation of the tangent line to the graph of f at the indicated point
    f(x)=x2+1 at (2, 5)
    Here is the setup:
    \lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1  - 5}}{{\Delta x}}
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by Plato View Post
    Here is the setup:
    \lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1  - 5}}{{\Delta x}}
    Hmm, so I see you used 2 as the x.
    This would simplify it down to (4△x+△x2)/△x
    Which then turns into 4x+△x?

    I'm not sure if I did it right, the answer at the back of the book is 4x-3.
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by skeeter View Post
    f(x) = x^2+1

    f(x + \Delta x) = (x+\Delta x)^2 + 1 (you forgot the 1)

    try again ...
    Um, I thought I added the +1 o.o
    I used the lim △x->0 (f(x+△x)-f(x))/△x, where this is the definition of a derivative.
    Could you explain more on where the +1 goes?

    Thanks
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by Plato View Post
    Here is the setup:
    \lim _{\Delta x \to 0} \frac{{f(2 + \Delta x) - f(2)}}{{\Delta x}} = \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1  - 5}}{{\Delta x}}
    \lim _{\Delta x \to 0} \frac{{4 + 4\Delta x + \left( {\Delta x} \right)^2+1  - 5}}{{\Delta x}}=\lim _{\Delta x \to 0} \frac{4\Delta x + \left( {\Delta x} \right)^2}{{\Delta x}}=\lim _{\Delta x \to 0} {4 + \left( {\Delta x} \right)}
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  7. #7
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by Chaim View Post
    Hmm, so I see you used 2 as the x.
    This would simplify it down to (4△x+△x2)/△x
    Which then turns into 4x+△x?
    No, it doesn't: 4△x/△x= 4, not "4x". Now take the limit as △x goes to 0. That gives the derivative at x= 2 and so the slope of the tangent line at x= 2.

    I'm not sure if I did it right, the answer at the back of the book is 4x-3.
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    Re: Finding the Equation of Tangent Line

    Quote Originally Posted by Chaim View Post
    Um, I thought I added the +1 o.o
    I used the lim △x->0 (f(x+△x)-f(x))/△x, where this is the definition of a derivative.
    Could you explain more on where the +1 goes?

    Thanks
    Because f(x)= x^2+ 1, it goes in both f(x+△x) and f(x). You added it in one but not the other:
    f(x)= x^2+ 1 so that f(x+\Delta x)= (x+\Deltas x)^2+ 1= x^2+ 2x\Delta x+ (\Delta x)^2+ 1 Then when you subtract the two, the two "1"s will cancel.
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