Exercise 6.2 · Q2: Show that $e^{2x}$ is increasing on $\R$

Basics you need

Solution

1. Differentiate with the chain rule

   Write $f(x) = e^{t}$ with the inner function $t = 2x$ and the outer function $e^{t}$⁠. Then

   $$\begin{aligned} f'(x) &= \frac{d}{dt}\left(e^{t}\right)\cdot\frac{dt}{dx}\\ &= e^{t}\cdot 2 = 2e^{2x}. \end{aligned}$$

   This holds at every real $x$⁠: $e^{2x}$ is differentiable everywhere (so also continuous everywhere).

   The derivative of $e^t$ is $e^t$ only when we differentiate with respect to its own input $t$⁠. Our input is $2x$⁠, which changes twice as fast as $x$⁠, so the chain rule brings in the factor $\frac{dt}{dx} = 2$⁠. Forgetting it gives $e^{2x}$⁠, which happens to be positive too, but is the wrong derivative.
   Chain rule (Theorem 4, Section 5.3.1): $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}$⁠. Basics: the chain rule

2. The derivative is positive everywhere

   For every real $x$⁠, $2x$ is a real number, so $e^{2x} > 0$⁠. Multiplying by the positive number 2 keeps it positive:

   $$f'(x) = 2e^{2x} > 0 \quad\text{for every } x \in \R.$$
   This is the whole heart of the problem. There is no point where $f'$ is zero or negative, so there is nothing to split up: no critical points, one interval, one sign.
   $e^t > 0$ for all real $t$ (Basic 3), and a product of two positive numbers is positive. Basics: the sign of a product

3. Apply Theorem 1 between any two inputs

   Let $x_1 < x_2$ be any two real numbers. On $[x_1, x_2]$ the function $f$ is continuous, it is differentiable on $(x_1, x_2)$⁠, and $f'(x) > 0$ there (Step 2). By Theorem 1 (a), $f$ is increasing on $[x_1, x_2]$⁠. The book's proof gives more: by the Mean Value Theorem there is a $c$ between $x_1$ and $x_2$ with

   $$\begin{aligned} f(x_2) - f(x_1) &= f'(c)\,(x_2 - x_1)\\ &= 2e^{2c}(x_2 - x_1) > 0, \end{aligned}$$

   because both factors are positive. So $f(x_1) < f(x_2)$ whenever $x_1 < x_2$⁠.

   Theorem 1 is stated for a closed interval $[a, b]$⁠. The question is about all of $\R$⁠. Since every pair $x_1 < x_2$ sits inside its own interval $[x_1, x_2]$⁠, applying the theorem there for an arbitrary pair settles every pair at once.
   The Mean Value Theorem, used in the book's proof of Theorem 1: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$⁠, then $f(b) - f(a) = f'(c)(b - a)$ for some $c$ in $(a, b)$⁠. The book cites it as "Theorem 8 in Chapter 5", but this reprint's Chapter 5 does not contain it, so it is quoted here without proof. Lesson: increasing and decreasing functions

4. Conclude

   We have shown $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ for all real $x_1, x_2$⁠. So $f$ is strictly increasing on $\R$ (Definition 1 (iv)) and hence $f(x) = e^{2x}$ is increasing on $\R$.

   Strictly increasing gives $f(x_1) < f(x_2)$⁠, and "$<$" implies "$\le$", which is what Definition 1 (i) asks for.

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