Proof/informal argument that $\exp$ is increasing at an increasing rate without calculus

Question

We've

• informally defined $\ln$ as giving the area under $y=1/x$;
• defined $\exp$ as the inverse of $\ln$;
• shown that $\exp$ is (strictly) increasing.

Now, how might we also show that $\exp$ is increasing at a (strictly) increasing rate, but without using calculus?

(A rigorous proof is ideal but an informal argument would do too. Context: High school.)

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Let $f:D \to C$ be a real-valued function of a real variable. We say that

• $f$ is strictly increasing if for all $a,b \in D$ with $a<b$, we have$f(a)<f(b)$;
• moreover, $f$ is strictly increasing at a strictly increasing rate if for all $a,b,c \in D$ with $a<b$ and $c-b=b-a$, we have$f(b)-f(a)<f(c)-f(b)$.

Answer 1

Here is an argument without using any limits. If you don't want to define the reals, you can assume the domain consists of the rationals. We will assume without proof that given any positive continuous function $f$ on an interval $[a,b]$, that we can define $A_f(a,b)$ to be the area under the graph $y=f(x)$, $a \le x \le b$. We also assume the following:

1. If $f < g$ on the interval $[a,b]$, then $A_f(a,b) < A_g(a,b)$.
2. The area is invariant under sliding along the $x$ axis. In other words, if $a < b$, $\Delta> 0$, and $g(x) = f(x+\Delta)$, then $$ A_g(a,b) = A_f(a+\Delta,b+\delta). $$
3. Additivity: If $a \le b \le c$, then $$ A(a,c) = A(a,b) + A(b,c). $$ Denote the area under $1/t$, $a \le t \le b$ by simply $A(a,b)$ and Define $$ \ln x = A(1,x). $$

By additivity, if $x < y$, then $$ \ln y = A(1,y) = A(1,x) + A(x,y) > \ln x, $$ so $\ln x$ is an increasing function of $x$.

Moreover, given $\Delta > 0$, since $1/t$ is a decreasing function, if $a < b$, then by the invariance under sliding, $$ A(a,a+\Delta) < A(b,b+\Delta). $$ You can also see this geometrically simply by sliding the graph of $1/x$, $b \le x \le b+\Delta$ left to the interval $a \le x \le a+\Delta$ and observing that the region under this graph lies inside the graph of $1/x$, $a \le x \le a+\Delta$

Therefore, given any $\Delta > 0$ and $a < b$, $$ (\ln (b+\Delta) - \ln b) - (\ln (a +\Delta) - \ln a) < 0. $$ Therefore, the graph of $\ln x$ is increasing and concave down.

We now define $e^x$ to be the inverse function of $\ln x$. Its graph is the reflection of the graph of $\ln x$ about the line $y=x$. Therefore, its graph is increasing and concave up. In particular, if $a < b$ and $\Delta > 0$, then $$ (e^{b+\Delta}-e^b) - (e^{a+\Delta}-e^a) > 0. $$ It follows that its rate of increase is increasing.

Answer 2

Hint I

It is the case that $e^x \ge \frac{x^3}{6}+\frac{x^2}{2}+x+1$ for all values of $x \in \mathbb{R}$.

This result can be derived from the Taylor Expansion of $e^x$ (I'm not sure whether or not you have come across this concept before). This is lower bound is strictly increasing for all values of $x \in \mathbb{R}$.

Therefore, $e^x$ is greater than a strictly increasing function, which (at the very least) tells us that the second derivative of $e^x$ will grow indefinitely as $x \rightarrow \infty$.

If you already know about the taylor expansion of $e^x$ or are curious about this, then you can use this to build up an informal intuition of why it might be the case that $e^x$ must be strictly increasing.

Hint II

To give you an even more informal perspective, we can simply study the graph of $e^x$ and look at the intuition behind why it has a strictly positive second derivative.

Not only is the graph increasing (postive first derivative), but we can see (informally), that the speed at which it is increasing is increasing for larger values of $x$ (positive second derivative).