# Thread: Show that f only takes two values

1. ## Show that f only takes two values

Hi

Question:

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

Is this a correct/complete solution?
My solution:

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

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

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

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

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

2. Originally Posted by Twig
Hi

Question:

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

Is this a correct/complete solution?
My solution:

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

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

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

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

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

The range of $\displaystyle \arctan x$ is $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2} \right)$. I think this is correct. But if $\displaystyle f$ is constant on $\displaystyle \mathbb{R} \ \backslash \{0 \}$, wouldn't $\displaystyle f(x)= c$ for all $\displaystyle x$ in that domain, for some number $\displaystyle c$?

3. Originally Posted by Twig
Hi

Question:

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

Is this a correct/complete solution?
My solution:

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

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

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

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

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

You can verify your result directly:

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

Now plug in the given terms:

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

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

If x < 0 the numerator of the fraction becomes negativ too and $\displaystyle f(x) = -\dfrac{\pi}2$

4. Originally Posted by Sampras
The range of $\displaystyle \arctan x$ is $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2} \right)$. I think this is correct. But if $\displaystyle f$ is constant on $\displaystyle \mathbb{R} \ \backslash \{0 \}$, wouldn't $\displaystyle f(x)= c$ for all $\displaystyle x$ in that domain, for some number $\displaystyle c$?
But no one said that f was constant on $\displaystyle \mathbb{R} \ \backslash \{0 \}$. 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 $\displaystyle (-\infty, 0)$ and on $\displaystyle (0, \infty)$ but not necessarily the same constant on those two intervals.

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

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

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

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

$\displaystyle =\,0$

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

6. 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. )

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

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

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

$\displaystyle =\,0$

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

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