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Math Help - Show that f only takes two values

  1. #1
    Senior Member Twig's Avatar
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    Show that f only takes two values

    Hi

    Question:

    Show that f(x) = arctan(\frac{1}{x}) + arctan(x) \, , x\neq 0 only takes two values. Show this, and decide these values.


    Is this a correct/complete solution?
    My solution:

    f'(x) = -\frac{1}{1+x^{2}} + \frac{1}{1+x^{2}} = 0 , so  f is a constant function on  \mathbb{R} - [0]

    Then take any x<0 , say  x = -1 .

     f(-1) = -\frac{\pi}{2} . This is the only value f takes on the interval  x<0 since f is constant on this interval.

    In the same way, we show that for  x>0 ,  f = \frac{\pi}{2} .

    Thus the values are  -\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}

    Thx for your input
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Twig View Post
    Hi

    Question:

    Show that f(x) = arctan(\frac{1}{x}) + arctan(x) \, , x\neq 0 only takes two values. Show this, and decide these values.


    Is this a correct/complete solution?
    My solution:

    f'(x) = -\frac{1}{1+x^{2}} + \frac{1}{1+x^{2}} = 0 , so  f is a constant function on  \mathbb{R} - [0]

    Then take any x<0 , say  x = -1 .

     f(-1) = -\frac{\pi}{2} . This is the only value f takes on the interval  x<0 since f is constant on this interval.

    In the same way, we show that for  x>0 ,  f = \frac{\pi}{2} .

    Thus the values are  -\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}

    Thx for your input
    The range of  \arctan x is  \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) . I think this is correct. But if  f is constant on  \mathbb{R} \ \backslash \{0 \} , wouldn't  f(x)= c for all  x in that domain, for some number  c ?
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  3. #3
    Super Member
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    Quote Originally Posted by Twig View Post
    Hi

    Question:

    Show that f(x) = arctan(\frac{1}{x}) + arctan(x) \, , x\neq 0 only takes two values. Show this, and decide these values.


    Is this a correct/complete solution?
    My solution:

    f'(x) = -\frac{1}{1+x^{2}} + \frac{1}{1+x^{2}} = 0 , so  f is a constant function on  \mathbb{R} - [0]

    Then take any x<0 , say  x = -1 .

     f(-1) = -\frac{\pi}{2} . This is the only value f takes on the interval  x<0 since f is constant on this interval.

    In the same way, we show that for  x>0 ,  f = \frac{\pi}{2} .

    Thus the values are  -\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}

    Thx for your input

    Your result is OK.

    You can verify your result directly:

    \arctan(x_1)+\arctan(x_2)=\arctan\left(\dfrac{x_1+  x_2}{1-x_1 \cdot x_2}\right)

    Now plug in the given terms:

    f(x)=\arctan(x)+\arctan\left(\frac1x\right)=\arcta  n\left(\dfrac{x+\frac1x}{1-x \cdot \frac1x}\right)

    Since {\dfrac{x+\frac1x}{1-x \cdot \frac1x}}{ = +\infty} ~ \implies~ f(x) = +\dfrac{\pi}2

    If x < 0 the numerator of the fraction becomes negativ too and f(x) = -\dfrac{\pi}2
    Attached Thumbnails Attached Thumbnails Show that f only takes two values-arctansum.png  
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  4. #4
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    Quote Originally Posted by Sampras View Post
    The range of  \arctan x is  \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) . I think this is correct. But if  f is constant on  \mathbb{R} \ \backslash \{0 \} , wouldn't  f(x)= c for all  x in that domain, for some number  c ?
    But no one said that f was constant on  \mathbb{R} \ \backslash \{0 \} . What is true is that the derivative of f is 0 on that set. That tells us that f is constant on any interval on which it is differentiable. Thus, f is constant on (-\infty, 0) and on (0, \infty) but not necessarily the same constant on those two intervals.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Here is another solution using calculus. (Since this topic is started in the Calculus section, I presume it is okay to use calculus. )

    f(x)\,=\,\arctan\left(\frac1x\right)+\arctan x

    \implies\ f'(x)\,=\,\frac{-\frac1{x^2}}{1+\left(\frac1x\right)^2}+\frac1{1+x^  2}

    =\,\frac{-1}{1+x^2}+\frac1{1+x^2}

    =\,0

    Hence f is constant over any interval on which it is continuous. Since it is continuous over (0,\,\infty) and over (-\infty,\,0), determine the constant value of f in each of these two invervals. For example, for x>0, we have f(1)=\arctan1+\arctan1=\frac\pi2 and so f(x)=\frac\pi2 for all x>0. Do the same thing for the negative interval.
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  6. #6
    Senior Member Twig's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Here is another solution using calculus. (Since this topic is started in the Calculus section, I presume it is okay to use calculus. )

    f(x)\,=\,\arctan\left(\frac1x\right)+\arctan x

    \implies\ f'(x)\,=\,\frac{-\frac1{x^2}}{1+\left(\frac1x\right)^2}+\frac1{1+x^  2}

    =\,\frac{-1}{1+x^2}+\frac1{1+x^2}

    =\,0

    Hence f is constant over any interval on which it is continuous. Since it is continuous over (0,\,\infty) and over (-\infty,\,0), determine the constant value of f in each of these two invervals. For example, for x>0, we have f(1)=\arctan1+\arctan1=\frac\pi2 and so f(x)=\frac\pi2 for all x>0. Do the same thing for the negative interval.

    Thanks everyone for your input!

    TheAbstractionist, how is your solution different from mine?

    Maybe I also should have stated that  f is continous on  \mathbb{R}-[0]
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  7. #7
    MHF Contributor matheagle's Avatar
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    It looked the same me.
    The discontinuity threw some people.
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