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Math Help - Interpreting Limits

  1. #1
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    Interpreting Limits

    Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

    lim [(2+h)^2-4]/h
    h->0

    my attempt:
    I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

    I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Evaluate the limit of the difference quotient. Interpret the limit as the slope of the tangent to a curve at a specific point.

    lim [(2+h)^2-4]/h
    h->0

    my attempt:
    I need help with the secondi part. I found the derivative which in this case is M_t=4 to the graph y=x^2 at x=_______? I am not sure how to find x =? at the specific point.

    I tried to put M_t=4 into the original eqtn and solve for h but that didn't work. Please provide detailed steps to finishing this problem. Thank you!
    f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

    for your problem, note that f(x) = x^2

    the limit is f'(2)
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  3. #3
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    \frac{{\left( {2 + h} \right)^2  - 4}}<br />
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}<br />
{h} = 4 + h,\,h \ne 0
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  4. #4
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    Quote Originally Posted by Plato View Post
    \frac{{\left( {2 + h} \right)^2 - 4}}<br />
{h} = \frac{{\left( {4 + 4h + h^2 } \right) - 4}}<br />
{h} = 4 + h,\,h \ne 0

    the answer is: 4, slope of tangent to y=x^2 at x=2

    My question is: how do I get x=2?
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  5. #5
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    Quote Originally Posted by skeske1234 View Post
    the answer is: 4, slope of tangent to y=x^2 at x=2
    My question is: how do I get x=2?
    That question really do not make any sense. Does it?

    If the question is, "find the x value where y=x^2 has slope 4."
    You would look at the derivative: 2x=4.
    From which it i clear that x=2.
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  6. #6
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    So you have already found the slope, m=4. I think you want the tangent line now. Based on your work, you used the difference quotient and pugged x=2 in to find the slope at this x value. So, at x=2 the slope is 4. So you have an x value, you now need a y-value in order to talk about the tangent line. Well, you know the tangent line has to touch our orinignal graph, y=x^2, so to find the y-value, just plug in our x-value. If x=2, y=4. So the point our line in tangent to is (2,4) with a slope of 4. Using point-slope form, our tangent line is:
    y-4=4(x-2)
    or
    y=4x-4
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