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X and y intercept

  1. May 11, 2006 #1
    Hi,
    I have to find x and y intercept of this function

    y= cos(e^(-x) + 2)

    This is what i have done so far, to find the x intercept i put y= 0

    0= cos(e^(-x) + 2)

    can i use Cos (A+B)= CosA CosB - SinA SinB here
    if i use that then i get

    0= cos e^(-x) cos2 - sin e^(-x) sin2

    i dont know what to do next. Please help, thanks.
     
    Last edited: May 11, 2006
  2. jcsd
  3. May 11, 2006 #2

    Integral

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    Where is the cos function zero?

    The argment of the cos must be that which makes cos(a)=0.
     
  4. May 11, 2006 #3
    cos function is zero at 90 degrees. How does that help.
     
  5. May 11, 2006 #4
    1) I believe you will need to use radians to solve this problem, because it will not really make sense to carry out algebraic manipulations using an angle in degrees.

    2) To find the x-intercept, we need to solve the equation [tex] cos(e^{-x} + 2)\ =\ 0[/tex]. First consider the equation [tex] cos(y)\ =\ 0 [/tex], then see how this can be used to solve the more complicated equation. I don't think the compound angle formula is really needed in this question.

    3) Does the question give you a range of x values for which the function is defined? If not, how many x-intercepts can there be? Also, it may be helpful to take note of the range of values [tex]e^{-x}[/tex] can take.
     
    Last edited: May 11, 2006
  6. May 12, 2006 #5

    VietDao29

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    Have you found out the y-intercept (which is pretty easier) yet?
    ------------
    In most calculus problems, the angle is in radians, so we don't really use degrees here.
    [tex]\cos x = 0 \Leftrightarrow x = \frac{\pi}{2} + k \pi, \ k \in \mathbb{Z}[/tex]
    Now, just take one minute to look at the unit circle to see if you can get it.
    Now if [tex]\cos (e ^ {-x} + 2) = 0[/tex]
    What should e-x + 2 equal?
    Can you go from here? :)
     
  7. May 12, 2006 #6
    Thanks pizzasky and VietDao29. No range of x values is given.

    [tex] (e ^ {-x} + 2) = \frac{\pi}{2}[/tex]

    [tex] (e ^ {-x} ) = \frac{\pi}{2} - 2[/tex]

    [tex] {-x} = \log \frac{\pi}{2} - 2[/tex]

    So if i substitute the value of -x in the equation i get

    cos(pi/2) = 0
     
    Last edited: May 12, 2006
  8. May 12, 2006 #7

    HallsofIvy

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    However, cosine is periodic. The graph of y= cos(x) crosses the x-axis an infinite number of times and so has an infinite number "x-intercepts".
    If no range was given for x, [itex]cos(e^{-x}+2)= 0[/itex] when [itex]e^{-x}+ 2= 2n\pi+ \frac{\pi}{2}[/itex] or [itex]e^{-x}+2= 2n\pi- \frac{\pi}{2}[/itex] where n can be any integer.
     
  9. May 12, 2006 #8
    I do not quite agree with your statement " [tex] -x = \log \frac{\pi}{2} - 2[/tex]". Shouldn't the natural logarithm be used, instead of "log"?

    Also, since no range of x values is given, how many x-intercepts can there be? Take a look at VietDao29's post for a hint!
     
  10. May 12, 2006 #9

    HallsofIvy

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    Many "advanced" text (beyond calculus) use "log x" to mean the natural logarithm and don't mention the common logarithm.

     
  11. May 12, 2006 #10
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    Oops, I did not know that...:blushing: But shouldn't it be [tex] -x = \log (\frac{\pi}{2} - 2)[/tex] (which, incidentally, is undefined)?
     
  12. May 13, 2006 #11
    Thanks everyone, so there are finite number of x intercepts since no range is given.
     
  13. May 14, 2006 #12
    You mean "infinite" number of x intercepts? Whilst there is no limit to the number of x-intercepts, some of the possible intercepts must be rejected. For instance, 1 of the intercepts is calculated to be [tex] x = -\log (\frac{\pi}{2} - 2)[/tex]. But is this possible?
     
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