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Math Help - How Do I Find the x and y intercepts??

  1. #1
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    How Do I Find the x and y intercepts??

    f(x)=xln(x^2)
    Thanks in advance
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  2. #2
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    To find the x intercept, solve for f(x) = 0
    To find the y intercept solve  y = xln(x^2) when x = 0

    Note that there may not necessarily be an x or y intercept
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  3. #3
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    Quote Originally Posted by redsox25 View Post
    f(x)=xln(x^2)
    Thanks in advance
    Since ln(x^2)= 2ln(x) that is the same as  f(x)= 2xln(x). There is no x intercept because ln(0) is not defined. To find the y intercept, use the fact that ln(1)= 0. f(1)= 2(1)ln(1)= 0. If x> 1, 2x ln(x)> 0 and if x< 1, 2xln(x)< 0 so that is the only y-intercept.
    Last edited by mr fantastic; October 12th 2009 at 06:53 PM. Reason: Fixed a latex tag
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    so when f(x)=0

    it's 0=xln(x^2)

    Therefore 0=x and 0= \ln (x^2)

    0= \ln(x^2)

    \Rightarrow e^0 = e^{\ln (x^2)}

    \Rightarrow 1=x^2

    \Rightarrow x=-1, 1

    but when i graph it x=0 doesnt exist! why is this??
    Last edited by mr fantastic; October 12th 2009 at 06:57 PM. Reason: Improved formatting for better readability
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  5. #5
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    To find the y intercept, use the fact that ln(1)= 0. f(1)= 2(1)ln(1)= 0. If x> 1, 2x ln(x)> 0 and if x< 1, 2xln(x)< 0 so that is the only y-intercept.
    While it is true that ln(x) > 0 and ln(x) = 2ln(x), the question did ask for the intercept of ln(x), not 2ln(x). There is a difference, for example  y = \frac{x^2}{x} has no y intercept but if you simplify it to y = x it does have one.

    In which case we also need to consider the negative values of x because it is squared to become positive within the natural logarithm function.

    but when i graph it x=0 doesnt exist! why is this??
    This is because ln(0) is undefined. the point (0,0) becomes a point of discontinuity. When graphing, you draw a hollow circle around the point to indicate this.
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  6. #6
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    Quote Originally Posted by redsox25 View Post
    so when f(x)=0

    it's 0=xln(x^2)

    Therefore 0=x and 0= \ln (x^2)

    0= \ln(x^2)

    \Rightarrow e^0 = e^{\ln (x^2)}

    \Rightarrow 1=x^2

    \Rightarrow x=-1, 1

    but when i graph it x=0 doesnt exist! why is this??
    The limiting value of your function as x --> 0 is 0 but it's not defined at x = 0 for reasons given earlier. The graph will have a 'hole' at (0, 0).
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