Why is a translated exponential function considered an exponential function?

Question

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?

Xander Henderson · Accepted Answer

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.

Sue VanHattum · Answer

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.

Answered by Sue VanHattum on September 6, 2021

Pedro · Answer

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).

Nick C · Answer

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.]

Why is a translated exponential function considered an exponential function?

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