Why is a translated exponential function considered an exponential function?

I am tutoring a student preparing to take Calculus 1 at a university. This student hasn't taken precalculus for a year, so I have been drilling him on definitions, rules, and theorems from a college level algebra course and precalculus. We were discussing types of functions. The following problem was brought up on an online quiz:

Classify the following function:

$$f(z)=5e^z+3$$

Now, I've visited many sites and they all seem to conclude that the following is the definition of an exponential function:

$$f(x)=ab^x \qquad \text{or} \qquad f(x)=ab^{cx+d}$$

with suitable restrictions on constants $$a,b,c,d$$.

So why isn't this function $$f(z)$$ above considered an exponential function? Certainly the $$3$$ represents only a shift of the exponential function up by three units. When these types of shifts are applied to polynomials, rational functions, trigonometric functions, they are still considered of that type. So why the change for exponential function? Why would a vertical shift be excluded in the definition?

• First, the extra $b^d$ term contributes no added generality. As to the main question, to me the key property is that $\exp \Sigma = \Pi \exp$, which is why I'd incline to refer to your example as just "(scaled) exponential plus/up to a constant". Jun 26 '20 at 15:24
• Okay. But why not have the definition include the translations; i.e., $$f(x) = ab^{(x-h)}+k$$ where $h$ and $k$ represent the horizontal and vertical translations respectively? Jun 26 '20 at 15:31
• This seems like a question about definitions, not a question about education, so it would be more appropriate for math.SE.
– user507
Jun 26 '20 at 18:07
• @BenCrowell I don't think that questions about definitions are necessarily off-topic here. It is one thing to ask about the nature of a definition from the point of view of reading an writing mathematics, and another to ask about definitions from the point of view of teaching. That the question was asked here adds the implicit question "and how should I teach this to a student?" Jun 26 '20 at 18:10
• It may also be worth noting that these kinds of transformations can change the way in which a function is classified. For example, a sinusoidal function is typically defined to be a function of the form $x \mapsto A \sin(\omega x + \varphi)$. Translating it up gives something which is not sinusoidal. A linear mapping is a function which satisfies $f(x+y) = f(x) + f(y)$ and $f(cx) = cf(x)$; the only functions are $\mathbb{R}$ which do this are of the form $x \mapsto ax$. Translations of the form $x \mapsto ax + b$ are affine, not linear, in this context. Jun 26 '20 at 19:58

To start with an opinion, I think that this classification exercise is kind of silly. The student is being asked to put functions into some categories without having a clear idea about what those categories mean or are used for. We introduce definitions and categorizations in order to help us understand abstract ideas. A definition without the underlying motivation is quite hard to grasp. As such, this is an exercise in rote regurgitation—I fail to see its value. :\

That being said, I would argue that there is (up to scaling by a constant) only one exponential function: $$x \mapsto \mathrm{e}^{x}.$$ This function is the unique solution to the initial value problem $$\begin{cases} u' = u \\ u(0) = 1. \end{cases}$$ The defining characteristic of the natural exponential function is that it is its own derivative. More generally, we have $$b^x = \mathrm{e}^{\log(b) x} \implies \frac{\mathrm{d}}{\mathrm{d}x} b^x = \log(b) \mathrm{e}^{\log(b) x} = \log(b) b^x.$$ Thus a function $$x \mapsto b^x$$ has the property that it is proportional to its own derivative. This is what I take to be the defining characteristic of an exponential function. That is, the rate at which an exponential function changes is equal to (or, at least, proportional to) the value of the function.

Any function of the form $$x \mapsto a \mathrm{e}^{bx} \tag{1}$$ has this property. Functions of the form $$x \mapsto a \mathrm{e}^{bx} + k \tag{2}$$ do not have this property. Because this property is important, it is reasonable to classify these two types of functions differently. Functions of type (1) are exponential, and functions of type (2) are not.

The problem here is that exponential functions are transcendental. You can't really discuss transcendental functions without relying on concepts from analysis (limits, continuity, differentiability, etc). As such, the most important feature of an exponential function (it is proportional to its own derivative) is inaccessible to a student who has not taken calculus. Of course, this renders the question "is this an exponential function or not?" completely mysterious to a precalculus student.

• I think you have a really good defense of the standard definition here. I would suggest you rethink "A definition without the underlying motivation is quite hard to grasp. As such, this is an exercise in rote regurgitation". It's clearly not literally rote regurgitation; it's a pattern-matching exercise, which is something different. I think exercises like these are actually pretty good, much too often overlooked, tests of whether a student has studied and learned a definition or not. I'm not sure how else to directly test whether someone really knows a definition, in fact. Jun 29 '20 at 6:50

The working definition I have in my head doesn't fit the more rigorous definitions others have put in their answers. I think of exponential growth and decay as being constant percentage growth or decay from or toward an asymptote. My favorite example is temperature of an object, which is shifted with the ambient temperature being the asymptote. I use y = a*b^x + c.

• I agree with you that exponential growth and decay are modeled by functions of the form $ab^x + c$, but I am not sure that this immediately implies that exponential growth is modeled by an exponential function. As Pedro suggested, the term-of-art would be, I think, "a function of exponential order". I am also not certain that I understand your statement that "exponential growth and decay [is] constant percentage growth or decay from or toward an asymptote". If it is a constant percentage, then $f' \propto f$; functions which decay to an asymptote don't have that property. Jun 26 '20 at 19:13
• In any event, I accept that there may be contexts in which a function $x \mapsto ab^x + c$ might be called an "exponential function", but I will stand by my assertion that the question is inappropriate in the context of a precalculus class (since, even in your heuristic definition, the derivative plays a key role). Jun 26 '20 at 19:15
• Newton's law of cooling (ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html) is usually stated in a way that makes the function look the way you and others here want, with temperature measured as a difference from the ambient temp. But when I work with precalc students, trying to get them to build this themselves as much as possible, I prefer y = a*b^x +c to that. Jun 27 '20 at 21:58
• It is a constant percentage of the distance from the asymptote. Jun 27 '20 at 21:58
• "It is a constant percentage of the distance from the asymptote." In that case, the function being described is not really a function of an independent variable $T$ (for temperature), but a function of the variable $T-T_A$, where $T_A$ is the ambient temperature. Cooling / warming is then an exponential function of this difference. Jun 27 '20 at 22:30

I say the key descriptor of a exponential function is constant multiplicative rate of change, much as the descriptor of a linear function is constant additive rate of change.

The function $$f(x)=a(1.5)^x$$ increases by 50% when $$x$$ increases by 1:

$$\frac{f(x+1)}{f(x)} = \frac{a(1.5)^{x+1}}{a(1.5)^x} = 1.5$$

But adding a non-zero constant changes that:

$$\frac{f(x+1)}{f(x)} = \frac{a(1.5)^{x+1}+c}{a(1.5)^x+c} \neq 1.5$$

So, if you define an exponential function by "constant multiplicative/percent rate of change", then you can't shift it.

[This is how I would explain it to an algebra student. I think the derivative argument would be great for a calculus student.]

Now, I've visited many sites and they all seem to conclude that the following is the definition of an exponential function: $$f(x)=ab^x$$, $$f(x)=ab^{cx+d}$$ with suitable restrictions on constants $$a,b,c,d$$.

These definitions are not good (unless the restrictions are $$a=1$$ in the first case and $$ab^d=1$$ in the second). A reasonable definition of "exponential function" should imply that it satisfies the basic rule of exponents $$a^na^m=a^{n+m}$$. That is, for an exponential function $$f$$ the property $$f(x+y)=f(x)f(y)\tag{1}$$ should be valid because it is this property that characterize the concept of "exponential". Translated exponential functions should not be considered exponential functions due to the same reason.

Remark 1: Under suitable hypotheses, it is possible to prove that the only functions that satisfy $$(1)$$ have the form $$f(x)=a^x$$ (with $$a=f(1))$$.

Remark 2: Usually, a definition is a matter of taste. Therefore, it is not wrong to define anything you want as "exponential function". Probably, it will only be unusual and not convenient.

Edit.

Remark 3: In science and engineering, functions that "behave" as exponential functions as all types mentioned in this post are usually called functions of "exponential order" (however, the concept of "exponential order" includes many other types of functions).

• I don't think this answer accurately represents the usage of the term by knowledgeable writers. $2e^x$ is an exponential function. In science and engineering, $a$ typically has units, so it doesn't make sense to introduce a constraint that $a=1$.
– user507
Jun 26 '20 at 18:07
• @BenCrowell There a are many knowledgeable writers who define exponential function as $a^x$ and call $2e^x$ a function of exponential order. Jun 26 '20 at 18:18
• @Pedro: Did you really mean to write "of exponential order"? Constant functions, polynomials, sums of exponentials and polynomials, etc. are all of exponential order (a well known phrase used in connection Laplace transforms, in complex analysis, and in a few other areas). Jun 26 '20 at 19:16
• @DaveLRenfro Thanks for the correction. I corrected the post. Jun 26 '20 at 19:27