# Let $f(x) = \frac{1-e^{2x}}{x}$. Solve $f(x)-2f(\frac{x}{2}) = 0$ in the interval $]-\infty,0[$

Let $f$ be a function of domain $\mathbb{R}$ defined by $$f(x)= \left\{ \begin{array}{lcc} \frac{1-e^{2x}}{x} & x < 0 \\ \\ -2 & x = 0 \\ \\ \frac{\ln(x^3)-x}{2x}, & x> 0 \\ \\ \end{array} \right.$$

Solve $f(x)-2f(\frac{x}{2}) = 0$ in the interval $]-\infty,0[$.

First I tried simplifying:

$$f(x)-2f(\frac{x}{2}) = 0\\ \Leftrightarrow \frac{1-e^{2x}}{x}-2\cdot\frac{1-e^{x}}{\frac{x}{2}} = 0\\ \Leftrightarrow \frac{1-e^{2x}}{x} = 2\cdot\frac{1-e^{x}}{\frac{x}{2}}\\ \Leftrightarrow \frac{1-e^{2x}}{x} = 4 \cdot \frac{1-e^{x}}{x} \\ \Leftrightarrow \frac{ \frac{1-e^{2x}}{x}}{ \frac{1-e^{x}}{x}} = 4 \\ \Leftrightarrow \frac{1-e^{2x}}{1-e^x}= 4$$

I wasn't entirely sure how to continue, so I tried first replacing x with 0 and since I got $\frac{0}{0}$, I decided to try calculating the limits:

$$\lim_{x \rightarrow 0} \frac{1-e^{2x}}{1-e^x} = \\ \frac{1-e^{2x}}{1-e^x} \cdot \frac{2x}{2x} \cdot \frac{x}{x} = \\ \frac{1-e^{2x}}{2x} \cdot \frac{x}{1-e^x}\cdot \frac{2x}{x} = \\ -\frac{e^{2x}-1}{2x} \cdot (-\frac{e^x-1}{x})^{-1}\cdot 2 = \\ -1 \cdot -1 \cdot 2 = 2$$

$$\lim_{x \rightarrow - \infty} = \frac{1-e^{2x}}{1-e^x} = 1$$

Since 4 is not in this range, does that mean this equation is impossible to solve?

My book says it is impossible, but doesn't explain why. Is my answer why this equation is impossible to solve?

By the way, I can't use L'Hopital's rule to solve limits.

Hint: $$\frac{1-e^{2x}}{1-e^x} = 1+e^x$$
• @MarkRead $1-e^{2x} = (1-e^x)(1+e^x)$ by difference of squares. – Zain Patel Mar 25 '17 at 1:38
• Right! I totally forgot about that. That way $1+e^x = 4 \Leftrightarrow e^x = 3 \Leftrightarrow x = \ln(3)$ and because $\ln(3) > 0$, there are no solutions in the wanted interval. Thanks. – Mark Read Mar 25 '17 at 1:41
$$f(x) = \frac{1-e^{2x}}{x}\implies f(x)-2f\left(\frac{x}{2}\right) = \frac{1-e^{2 x}}{x}-\frac{4 \left(1-e^x\right)}{x}=-\frac{\left(e^x-3\right) \left(e^x-1\right)}{x}$$