Exponential Functions and Graphs

Adding versus multiplying

Put two machines side by side. Both start at $3$. The first adds $4$ every time the input goes up by one; the second multiplies by $2$ every time the input goes up by one.

The adding row is the linear function $L(x) = 3 + 4x$. Its differences are constant: $7 - 3 = 4$, $11 - 7 = 4$, $15 - 11 = 4$. Its ratios are not: $\tfrac{7}{3}$, then $\tfrac{11}{7}$, then $\tfrac{15}{11}$, all different.

The multiplying row is the function $E(x) = 3 \cdot 2^x$. Now the tables trade places. Its differences are not constant ($3$, then $6$, then $12$), but its ratios are: $\tfrac{6}{3} = 2$, $\tfrac{12}{6} = 2$, $\tfrac{24}{12} = 2$.

That is the whole idea, and it deserves a name.

An exponential function is a function of the form

where the variable sits in the exponent and the base $b$ is a fixed number. (The restrictions on $b$ look arbitrary right now. Two sections from now you will see that they are forced.) Compare it with $x^2$ or $x^{1/3}$, where the variable sits in the base and the exponent is fixed. Those are power functions, and they behave nothing like this.

Check the ratio for the general rule. Step the input up by one:

The $x$ cancels completely, so the answer does not depend on where you started. Take a step of size $s$ instead and the same cancellation gives $\tfrac{f(x+s)}{f(x)} = b^{s}$. Equal steps in the input multiply the output by the same factor, everywhere along the curve. That property is the reason a bank account, a bacterial culture, and the gap between a cooling cup of coffee and the room around it all end up with the same kind of formula. That same property is the one we are going to take as the definition.

It also gives you a test with teeth. Faced with a table of equally spaced inputs, divide neighboring outputs rather than subtracting them. If every quotient is the same positive number other than $1$, the data fits an exponential rule, and that quotient is $b^{s}$, where $s$ is the spacing between the inputs. Only when the inputs step by one is the quotient the base itself, and the difference matters. If the inputs run $0, 2, 4$ and the outputs are $3, 12, 48$, the constant quotient $4$ is $b^{2}$, so the base is $2$, not $4$. If instead every difference is the same, the data is linear.

The two excluded quotients are worth naming. A constant quotient of $1$ repeats a single output forever, which is the constant function. And a constant negative quotient, as in $1, -2, 4, -8$, flips the sign at every step, which no positive base can do.

Equal ratios force the law of exponents

So far $b^x$ is a formula we wrote down. Now let us do it the other way around, and demand the behavior we want before we know any formula at all.

Suppose $f$ is a function defined for every real number $x$, and we insist on exactly three things.

1. It starts at $1$, so $f(0) = 1$. (A general starting value $a$ can be multiplied back in later.)
2. A step multiplies. For each step size $s$ there is some factor $m(s)$, depending on the step but never on where you started, with $f(x + s) = m(s) \cdot f(x)$ for every input $x$.
3. No jumps, and no doubling back. The graph is one unbroken curve, so nudging the input a little only nudges the output a little. And when the base turns out to exceed $1$, a bigger input always gives a bigger output.

That is all. We have not said the word exponent. The first two demands already pin down the law that governs everything that follows, and they will carry us all the way through the fractions. The third demand looks too obvious to be worth writing down, and we will spend it exactly once, at the very end, where it turns out to be indispensable. Watch for it.

One more piece of notation and we are ready to build. The function’s value at $x = 1$ is the factor produced by a one-unit step, so it deserves the name base:

The next section shows that $b$ is not just an important value. It is the only input you get to choose. Every other value of $f$ is then forced.

Building the function one input at a time

We know two things about $f$: its value at $0$ is $1$, and $f(x+y) = f(x) f(y)$. That is barely two sentences of information, but pinning down every value of $f$ from it, one kind of input at a time, is worth seeing in full, and the complete argument lives in Go Deeper below. What survives it, and what you will use for the rest of this course, are five facts.

• $f(x) > 0$ for every real $x$. Cutting any input in half writes $f(x)$ as a square, $f(x) = \left[f(x/2)\right]^{2}$, and a square is never negative; $f$ cannot be $0$ anywhere either, since that would force it to be $0$ everywhere, contradicting $f(0) = 1$.
• $f(n) = b^{n}$ for a positive integer $n$, from applying the law $n - 1$ times: $f(2) = f(1)f(1) = b^2$, $f(3) = f(2)f(1) = b^3$, and so on.
• $b^{0} = 1$ is forced, not chosen. Setting $y = 0$ in the law gives $f(x) = f(x) \cdot f(0)$, and dividing by $f(x) \neq 0$ leaves $f(0) = 1$.
• $b^{-n} = \dfrac{1}{b^{n}}$. Since $x$ and $-x$ add to $0$, the law gives $f(x) f(-x) = f(0) = 1$.
• $b^{p/q} = \sqrt[q]{b^{\,p}}$. Stacking a $\tfrac1q$-step $q$ times must return $b$, so $f\!\left(\tfrac1q\right)$ is the positive $q$-th root of $b$, and $p$ of those steps give the rest.

The domain is every real number, not only the rationals. The rationals are dense, so at this point the graph is already a fine dust of points that only looks like a curve. The third demand, that the graph has no jumps, is what fuses that dust into one unbroken curve: it forces the value at an irrational input like $\sqrt2$ to sit exactly where climbing rational powers on both sides squeeze it.

Why the base must be positive and never one

The restrictions $b > 0$ and $b \neq 1$ were stated without justification earlier. Here is why each one is unavoidable.

A negative base breaks immediately. Suppose someone hands you $b = -4$ and asks for the function $f(x) = (-4)^{x}$. At whole-number inputs nothing looks wrong: $f(2) = 16$ and $f(3) = -64$. But ask for $f\!\left(\tfrac12\right)$. Squaring it would have to give $f(1) = -4$, and no real number squares to a negative. So the function has no value at $x = \tfrac12$. The same failure repeats at $x = \tfrac14$, at $x = \tfrac16$, and at every fraction whose denominator is even once the fraction is written in lowest terms. So the graph is riddled with holes, and no unbroken curve exists to draw. (The lowest-terms wording is doing real work. Writing $\tfrac12$ as $\tfrac24$ does not rescue anything, but $\tfrac22$ is just the whole number $1$ in disguise, and $(-4)^{1}$ is perfectly fine.) A negative base is not merely inconvenient. It makes the function impossible.

A base of zero breaks too. With $b = 0$ every positive input gives $f(x) = 0$, which contradicts what we proved: an $f$ obeying the law of exponents with $f(0) = 1$ is never zero. And $f(-1)$ would ask for $\tfrac{1}{0}$.

A base of $1$ is legal but empty. The function $f(x) = 1^{x}$ does satisfy $f(x+y) = f(x)f(y)$, but it is the constant function $1$, a horizontal line. It never grows, never decays, takes the same output for every input, and so has no inverse to undo it. Nothing exponential happens, so we exclude it by convention rather than by necessity.

That leaves $b > 0$ with $b \neq 1$, which splits into exactly two behaviors: $b > 1$ and $0 < b < 1$.

The graph and what it tells you

Plot $y = 2^{x}$ and $y = \left(\tfrac12\right)^{x}$ on the same axes. They are not two unrelated shapes. Since $\left(\tfrac12\right)^{x} = \left(2^{-1}\right)^{x} = 2^{-x}$, replacing $x$ by $-x$ turns one into the other, and replacing $x$ by $-x$ is precisely the reflection across the $y$-axis you learned in the transformations lesson.

Read the picture against everything proved so far. Every statement in this list is about the parent curve $y = b^{x}$; the coefficient $a$ goes back in the next section, and it moves some of them.

• Domain. Every real number, because we built a value at integer, rational, and then irrational inputs.
• Range. All positive numbers. Nothing below the axis, since $f(x) = [f(x/2)]^2$ is never negative and never zero. And nothing positive is missed, because the curve is unbroken and climbs from values as close to $0$ as you like up to values as large as you like.
• The $y$-intercept is always $(0, 1)$, whatever the base, because $b^0 = 1$ for every $b$. The point $(1, b)$ is on the curve too, which is how you read the base off a graph.
• Horizontal asymptote $y = 0$. For $b > 1$ the curve flattens toward the $x$-axis on the left; for $0 < b < 1$ it does so on the right. It never arrives. This is a one-sided asymptote, unlike the two-sided approach you saw with rational functions.
• Monotonic. Under the standing assumptions ($b > 0$ and $b \neq 1$) the function is strictly increasing exactly when $b > 1$, and strictly decreasing exactly when $0 < b < 1$. There is no third case and no turning point anywhere.

That last bullet has a consequence worth its own display. A strictly increasing or strictly decreasing function never takes the same output twice, so an exponential function is one-to-one:

Read right to left it is obvious. Read left to right it is a tool: if you can write both sides of an equation as powers of the same base, you may simply equate the exponents.

Shifting and stretching

Everything you learned about transforming graphs applies here without change. The general transformed exponential is

and each constant does its usual job: $h$ shifts horizontally, $k$ shifts vertically, and $a$ stretches the curve vertically when $|a| > 1$ and compresses it when $0 < |a| < 1$ (flipping it upside down too when $a$ is negative). The one thing to watch is the asymptote. The parent curve hugs $y = 0$, so adding $k$ carries the asymptote with it, up to $y = k$. The range follows: all $y > k$ when $a > 0$, and all $y < k$ when $a < 0$.

There is one surprise, and it belongs to exponentials alone. Watch what a horizontal shift really does:

Since $b^{-h}$ is just a positive constant, sliding an exponential sideways is the same as stretching it vertically. Shifting $y = 2^{x}$ three units right multiplies it by $2^{-3} = \tfrac18$, so the “shifted” curve is the original squashed to one eighth of its height. No parabola behaves like that: sliding $y = x^2$ sideways genuinely moves its vertex. And the correspondence runs both ways, because the range of $b^x$ is all positive numbers, so every stretch factor $c > 0$ equals $b^{-h}$ for exactly one $h$. Naming that $h$ is a job we cannot yet do, and it is the reason the next lesson exists.

Nothing polynomial keeps up

A doubling function starts small. At $x = 3$, $2^{x} = 8$ while $x^{2} = 9$, so the humble parabola is ahead. It stays ahead for a while, and then it loses forever.

Why can the parabola never recover? Because of the ratio property, applied one step at a time. Induction proves it for every whole number: at $n = 4$ the two sides are equal, since $2^{4} = 16$ and $4^{2} = 16$, and once $2^{n} \geq n^{2}$ holds for one whole number it holds for the next one too.

The induction also shows why, and the reason is exactly the property this lesson started from. One step to the right multiplies $2^{x}$ by $2$, no matter where you are on the curve. The same step multiplies $x^{2}$ by $\left(\tfrac{n+1}{n}\right)^{2} = \left(1 + \tfrac{1}{n}\right)^{2}$, which is at most $\left(\tfrac54\right)^{2} = 1.5625$ once $n \geq 4$, and which keeps sliding toward $1$ as $n$ grows. A fixed factor of $2$ against a shrinking factor tending to $1$ can end only one way. That reasoning runs one whole-number step at a time, which is exactly what the induction proves; pinning the same fact down for every real $x$ past $4$, not only the whole numbers, needs calculus you have not met yet, though the picture above makes it easy to believe.

Nothing in that argument was special to the square. Against $x^{10}$ the exponential falls behind for a long time (at $x = 20$, $2^{x}$ is about a million while $x^{10}$ is about ten trillion). But the step factor of $x^{10}$ also decays toward $1$, so somewhere between $x = 58$ and $x = 59$ the exponential passes it and is gone. And the reason no polynomial can imitate an exponential in the first place is that constant ratios are not something a polynomial can produce: comparing leading terms shows the scale factor of any such polynomial would have to be exactly $1$, and only a constant function can have that.

Common mistakes