# Sketching $y = \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)$

The exercise is to graph this function below.

$$y = \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)$$

If the exercise were just one trigonometric function, then I would just put the expression inside arctangent in argtangent's domain and moreover put $$y$$ in the range of which the argtangent is defined. However there are two trigonometric functions in the equation above, and I do not know where I would start to able to sketch it.

I also want to add that I solved the equation algebraically by taking the tangent of both sides and using the subtaction of two angle for tangent

$$\tan(x-y)=\frac{\tan\ x + \tan\ y}{1 + \tan\ x*\tan\ y}$$

The equation simplified to $$\tan\ y = \sqrt3$$, which gives us $$y = \frac{\pi}{3}$$. Does this $$y$$ value help me sketch the function in anyway?

• The function is constant. A horizontal line. What you discovered is that $y=3$ regardless of which value of $x$ is chosen. Also, it is useful to ask yourself why $y$ isn't, for instance, $-2π/3$. And before wrapping up you might want to verify the domain of the function $y(x)$ is the real line. To answer your original question, though, my high school teacher used to say, "if you're not sure, take the first derivative," after all the derivative tells you all about the function $y(x)$ up to a constant. What is $\frac{d}{dx} y(x)$? – LPenguin Sep 26 '18 at 22:39
• How would I know that $y = \pi/3$ for all x in the domain and that the function is constant? And I think that this task is suppose to be solved without differentiating the function, considering the fact that we have not covered the differentiation chapter yet. – Bill Henson Sep 27 '18 at 11:25

## 1 Answer

Note that the arctan has an implied $$+k\pi$$ since $$\tan(x+k\pi) =\tan(x)$$, so you have to be careful. For example, if you naively evaluate $$\arctan(1)+\arctan(2)+\arctan(3)$$ you get $$\arctan(1)+\arctan(2) =\arctan(\frac{1+2}{1-1\cdot 2}) =\arctan(-3)$$ so $$\arctan(1)+\arctan(2)+\arctan(3) =\arctan(-3)+\arctan(3) =0$$, but the actual sum, using principal values, is $$\pi/2$$.

For this case, when $$x > 0$$,

$$\begin{array}\\ y &= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)\\ &= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}}\right)\\ &= t_1-t_2\\ \end{array}$$

Since $$e^x > 1$$, $$\frac{e^x-1}{\sqrt3} > 0$$ so $$0 < t_1 < \pi/2$$ and $$\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}} \gt \frac{1-4/3}{\sqrt3} =-\frac1{3\sqrt{3}}$$ so $$-\arctan(\frac1{3\sqrt{3}}) \le t_2 \lt \arctan(1/\sqrt{3}) =\pi/6$$.

Therefore $$-\pi/6 \lt t_1-t_2 \lt \pi/2-\arctan(\frac1{3\sqrt{3}}) \approx \pi/2-0.190$$ and, since $$\lim_{x \to \infty} t_1 = \pi/2$$ and $$\lim_{x \to \infty} t_2 = \pi/6$$, we have $$\lim_{x \to \infty} t_1-t_2 = \pi/3$$.

For $$x < 0$$ you can do a similar analysis using $$0 < e^x < 1$$.

• How can I use this to sketch the graph of the function? – Bill Henson Sep 27 '18 at 11:26
• The function is constant, – marty cohen Sep 27 '18 at 13:17
• How would we know that? – Bill Henson Sep 27 '18 at 14:34
• As OP wrote, using the subtraction formula for arctan gave $\arctan(\sqrt{3}) = \pi/3$. The only question is whether $\pm \pi/2$ needs to be added. – marty cohen Sep 27 '18 at 15:55