# Show that f only takes two values

• Jun 24th 2009, 11:23 AM
Twig
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 $-\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}$

• Jun 24th 2009, 12:12 PM
Sampras
Quote:

Originally Posted by Twig
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 $-\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}$

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 $c$?
• Jun 24th 2009, 12:17 PM
earboth
Quote:

Originally Posted by Twig
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 $-\frac{\pi}{2} \mbox{ and } \frac{\pi}{2}$

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 $f(x) = -\dfrac{\pi}2$
• Jun 24th 2009, 12:29 PM
HallsofIvy
Quote:

Originally Posted by Sampras
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.
• Jun 24th 2009, 12:30 PM
TheAbstractionist
Here is another solution using calculus. (Since this topic is started in the Calculus section, I presume it is okay to use calculus. (Evilgrin))

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

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

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

$=\,0$

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.
• Jun 25th 2009, 12:03 AM
Twig
Quote:

Originally Posted by TheAbstractionist
Here is another solution using calculus. (Since this topic is started in the Calculus section, I presume it is okay to use calculus. (Evilgrin))

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

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

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

$=\,0$

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.

Maybe I also should have stated that $f$ is continous on $\mathbb{R}-[0]$