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Math Help - Equation of tangent line

  1. #1
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    Equation of tangent line

    Hi!

    Is my solution correct?

    Let f(x) = x^2 -1. Find the equation of the line tangent to the graph of f at the point where x = 2.

    \lim_{h \to 0} = \frac{f(x+h) - f(x)}{h}

    \lim_{h \to 0} = \frac{f(2+h)^2 - f(2)}{h}

    \lim_{h \to 0} = \frac{\left[(2+h)^2 - 1\right] - (2^2-1)}{h}

    \lim_{h \to 0} = \frac{h^2 + 4h + 4 - 4 + 1}{h}

    \lim_{h \to 0} = \frac{h^2 + 4h + 1}{h}

    \lim_{h \to 0} = \frac{h(h+4) - 3}{h}

    \lim_{h \to 0} (h+4) = -3

    Limit at x=2 doesn't exist.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Equation of tangent line

    Quote Originally Posted by Unreal View Post
    \lim_{h \to 0} (h+4) = -3

    Limit at x=2 doesn't exist.
    That last line isn't correct and I'm not sure where it came from. Anyway...

    I wouldn't have done it quite this way...the usual method is to plug in x after you get the expression for the derivative.
    f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

    = \lim_{h \to 0} \frac{ ( (x + h)^2 - 1 ) - (x^2 - 1)}{h}

    etc.
    f'(x) = \lim_{h \to 0} \frac{2hx + h^2}{h} = \lim_{h \to 0}(2x + h) = 2x

    and so forth. Can you take it from here??

    -Dan
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  3. #3
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    Re: Equation of tangent line

    Quote Originally Posted by Unreal View Post
    Hi!

    Is my solution correct?

    Let f(x) = x^2 -1. Find the equation of the line tangent to the graph of f at the point where x = 2.

    \lim_{h \to 0} = \frac{f(x+h) - f(x)}{h}

    \lim_{h \to 0} = \frac{f(2+h)^2 - f(2)}{h}

    \lim_{h \to 0} = \frac{\left[(2+h)^2 - 1\right] - (2^2-1)}{h}

    \lim_{h \to 0} = \frac{h^2 + 4h + 4 - 4 + 1}{h}
    This line is wrong. Your basic problem appears to be arithmetic.

    \lim_{h \to 0} = \frac{h^2 + 4h + 1}{h}

    \lim_{h \to 0} = \frac{h(h+4) - 3}{h}

    \lim_{h \to 0} (h+4) = -3

    Limit at x=2 doesn't exist.
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  4. #4
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    Re: Equation of tangent line

    Equation:

    y = m(x-a) + f(a)

    y = 2(x-2) + 4

    OR

    y = 2x
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  5. #5
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    Re: Equation of tangent line

    Quote Originally Posted by Unreal View Post
    Equation:

    y = m(x-a) + f(a)

    y = 2(x-2) + 4

    OR

    y = 2x
    f'(x) = 2x, so f'(2) = 2(2) = 4, not 2.
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  6. #6
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    Re: Equation of tangent line

    New equation,

    y = 4(x-2) + 4

    OR

    y = 4x -4, 4(x-1)
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Equation of tangent line

    Quote Originally Posted by Unreal View Post
    New equation,

    y = 4(x-2) + 4

    OR

    y = 4x -4, 4(x-1)
    Still not going well. I don't think I've encountered that form for a line, or maybe I'm just not seeing it right. Try the slope-intercept form of a line:
    y = mx + b

    The slope for x = 2 is 2x = 2(2) = 4. So we have
    y = 4x + b

    You need to have the actual point where the tangent occurs, so what is f(2)? That will give you your point: (x, y) = (2, f(2)), and you plug that into your tangent line equation. That will give you the intercept for your tangent line.

    By the way, recalculate f(2). It's not 4.

    -Dan
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  8. #8
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    Re: Equation of tangent line

    y = 3 + 4(x-2)
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