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Math Help - Finding exact zeros and tangent lines.

  1. #1
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    Finding exact zeros and tangent lines.

    So, here's the problem:

    1) Let f be the function defined by f(x) = -A + ln(x^2), where A is the number of letters in your last name. (For me, that would be 6).
    (a) Find the exact zeros of f (using algebra, not your calculator).
    (b) Write an equation for the tangent line to the graph of f at x = -1.

    Can someone explain to me how to get started solving these questions?
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  2. #2
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    To begin with, x is a zero of f(x) when

    -6 + \ln x^2=0.

    Solving this equation gives us the zeros of f(x). To find the equation of the tangent line at x=-1, we differentiate f(x) to find its slope at x=-1, remembering that the equation for a line is

    y-y_1=m(x-x_1).
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Rumor View Post
    So, here's the problem:

    1) Let f be the function defined by f(x) = -A + ln(x^2), where A is the number of letters in your last name. (For me, that would be 6).
    (a) Find the exact zeros of f (using algebra, not your calculator).
    (b) Write an equation for the tangent line to the graph of f at x = -1.

    Can someone explain to me how to get started solving these questions?

    for a)

    -A + \ln (x^2) =0

    A = \ln (x^2)

    e^{A} = x^2

    x = \mp e^{\frac{A}{2}}


    for 2)

    f(x) = -A + \ln (x^2)

    f'(x) = 0 + \frac{2x}{x^2}

    f'(-1) = \frac{-2}{1} = -2 the slope of the function at x=-1

    the equation of the tangent line when x=-1 , first find y when x=-1

    f(-1) = -A + \ln (x^2) = -A + 0 =-A

    the equation

    y-(-A ) = slope (x-(-1))

    y+A = -2(x+1)
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  4. #4
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    Thank you to the both of you! That was really helpful.
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