How can I show that $|e^{2x} - e^{2y}| \leq 8|x - y|$ for all $x < y < \ln{2}$?

Question

This is what I have managed to accomplish:

We consider the function $f(t) = e^{2t}$. Then, as $f$ is continuous on the real number line, it is continuous on $[x, y]$.

By the Mean Value Theorem, $$ \exists c \in [x, y], \text{ such that } f'(c) = \frac{f(y) - f(x)}{y - x}. $$

That is, $$ \exists c \in [x, y], \text{ such that } 2e^{2c} = \frac{e^{2y} - e^{2x}}{y - x}. $$

As $$ \frac{e^{2y} - e^{2x}}{y - x} = \frac{e^{2x} - e^{2y}}{x - y}, $$ we have $$ \exists c \in [x, y], \text{ such that } 2e^{2c} = \frac{e^{2x} - e^{2y}}{x - y}. $$

This is where I do not know how to continue.

1. How do I incorporate the absolute value signs into this equation?
2. How do I even substitute $\ln{c}$ into the equation? The question states that $x < y < \ln{2}$, so with $c \in [x, y]$, it is the case that $c \neq \ln{2}$. Could there be an error in the question?

I would greatly appreciate if anyone were to stumble by my question.

Thank you so much.

Answer 1

You're almost there! As $c\mapsto 2e^{2c}$ is an increasing function, it attains it maximum on $(-\infty,\ln 2]$ at $\ln2$, meaning also that $2e^{2c}\leq2e^{2\ln 2}$ for all $c<\ln 2$. You thus have that

$$\left\lvert\frac{e^{2y}-e^{2x}}{y-x}\right\rvert=\lvert 2e^{2c}\rvert\leq 2e^{2\ln 2}=8.$$

Rearranging we get

$$\lvert e^{2x}-e^{2y}\rvert\leq 8\lvert x-y\rvert.$$