f(x+h) in the difference quotient

Question

When teaching students how to compute the difference quotient in a precalculus or calculus class, we need them to evaluate the expression

$$\frac{f(x+h) - f(x)}{h}$$

for various simple functions, like linear and quadratic functions. Let's say we are using the function $f(x) = x^2 + x + 1$ for example.

However, students find this to be more or less impossible. Specifically, when trying to find $f(x+h)$, students do all sorts of crazy things while computing the difference quotient. They replace $f(x+h)$ in the expression with things like:

• $f(x^2 + x + 1 + h)$
• $(x+h)^2 + x + 1$
• $x^2 + x + 1 + h$

which are all clearly wrong.

However, if you ask these same students this question:

Find $f(\text{Tomato Soup})$.

they are happy to do so and do it successfully.

In fact, my experience is that they can actually also complete this question:

Find $f(xxx)$.

This is also easy for them:

Find $f(y)$.

However, this question is a whole different story for them:

Find $f(x+h)$.

Why do students see $f(x+h)$ as fundamentally different from the others? What series of questions or conversations can be used to help them?

Answer 1

You have already applied some good diagnostic tests. I recommend the following additional diagnostics

What happens if you ask them to evaluate each of the following:

1. $f(3y)$: Passing this test indicates that they can think of a monomial as an input. I am guessing your students will pass this test given the other diagnostics you have run.
2. $f(3x)$: A student could pass the first test and fail this one if they don't think it is "kosher" to give $f$ an input of $3x$ when $f$ was defined by $f(x) = x^2 + x + 1$. They are experiencing variable scope anxiety. They are getting "the $x$ in $3x$" confused with "the $x$ in $f(x)$". To be more technical about it we should really say "The function $f: \mathbb{R} \to \mathbb{R}$ defined by $\forall x \in \mathbb{R}, f(x) = x^2+x+1$". The $x$ is bound by the universal quantifier.
3. $f(t+3)$: Passing this test indicates that they can think of a binomial as a "single input". A student who fails this test is not thinking of $(t+3)$ as a "single thing". They might also have difficulty distributing $(t+3)(a+b) = (t+3)a + (t+3)b$ since they cannot conceptualize the $(t+3)$ as a single number to be distributed.
4. $f(x+3)$: Again, a student could fail this test while passing the third test because of variable scope anxiety.
5. $f(j+h)$: A student could fail this test while passing all of the other tests because they think that a function should apply to only "one variable at a time". They view $x+3$ as a legit input since it is a single expression involving one variable. They freak out at $f(j+h)$ because they are not sure which variable they should be applying $f$ to. In essence they are seeing a function being applied to an expression with two variables and think that this will make $f$ a "two variable function" even though it is only a one variable function. They don't know what to do with the "second input".
6. $f(x+h)$ Again, a student could fail this test while passing test #5 because of variable scope confusion.

My success rate at diagnosing these issues is pretty good, but I have a poor cure rate...

Answer 2

I know a teacher who (at least in the past) would require students to write underlined blanks in place of the input whenever they were evaluating a function from its formula:

[Examples with $f(x)=5x^2-3x+1$]

Evaluate $f(6)$.

Answer:

$$\begin{align} f(6) &= 5(\text{_______})^2 - 3(\text{_______}) +1 \\ &= 5(6)^2 - 3(6) + 1\\ &= \dots\end{align}$$

Evaluate $f(3+h)$.

Answer:

$$\begin{align} f(3+h) &= 5(\text{_______})^2 - 3(\text{_______}) +1 \\ &= 5(3+h)^2 - 3(3+h) + 1\\ &= \dots\end{align}$$

This wouldn't stop a student from mishandling the $(3+h)^2$, but it might get the $h$ in the right place.

Answer 3

The issue here seems to be with substitution, a fundamental operation in mathematics.

Specifically, substituting $x$ with an expression that itself contains $x$.

The issue of how to perform substitution correctly actually arises in the lambda calculus and requires variable renaming to handle conflicts.

I propose using this idea of renaming to avoid confusion. Specifically, we "break down" the substitution $x \mapsto x + h$ into two steps: $x \mapsto y$ and $y \mapsto x + h$. Here, $y$ is what's called a fresh variable, that is, a variable that has not appeared before (and thus avoids any kind of conflict with the expressions we've working with so far). For example:

\begin{align} && f(x) &= x^2 + x + 1 \\ x &\mapsto y & f(y) &= y^2 + y + 1 \\ y &\mapsto x + h & f(x + h) &= (x + h)^2 + (x + h) + 1 \end{align}

Note: Make sure to emphasize that, when substituting a compound expression like $x + h$, it should be wrapped in parentheses, like $(x + h)$. This ensures that it's treated as a "single unit" in the underlying expression (e.g., avoids changing the order of operations).

In contrast, performing $x \mapsto x + h$ all at once induces an additional cognitive load. This cognitive load comes from having to keep track, as the expression is being written out, of which $x$s are "the original $x$s" versus the ones that come from the new expression. In particular, you have to make sure you don't miss an "original $x$" and make sure you don't accidentally replace a "new $x$". For example, consider the following thinking process:

\begin{align} & x^2 + x + 1 \\ & \text{We have to replace $x$ with $x + h$.} \\ & \text{Ok, let's start with the first occurrence of $x$.} \\ & (x + h)^2 + x + 1 \\ & \text{Hmm, which of the $x$'s am I supposed to replace now?} \end{align}

Answer 4

My solution is something like Nick's. I think it has helped my students, but I don't have hard data.

I scribble out the x's. So $f(x)=5x^2-3x+1$ becomes $f(scribble)=5scribble^2-3scribble+1$. And I talk about f of any mess gives us 5 times that mess squared minus 3 times the mess plus 1.

I then tell them to circle the x's. For f(x+h) we'll need an (x+h) inside each of those circles, and yes, we'll want those parentheses around it each time.

Then I give them a quiz on finding the derivative (and tangent line) from the definition. I let them take this quiz multiple times, hopefully until they get it right. Then they have to do it again on the first test.

Answer 5

I think the problem originates from conflicting mathematical notation. In elementary school multiplication is explicitly written out $a \times b$. By middle school it becomes just $ab$, so $a(b+c)$ means $a$ times $(b+c)$.

Then functions come in during the last few years of high school. And the notation there is similar to multiplication, and perhaps student thinks $f(x)$ should mean $f$ times $x$. So $f(x+h)$ should be $f$ times $(x+h)$.

One solution is to insist on explicit writing of all operations. So $a\times b$ for multiplication and perhaps $f \circ (x)$ or $f \& (x)$ for $f$ of/at $x$ or something similar. Of course the downside is lengthy writing. Probably $a\times b$ will be a hard sell, but I think explicit math symbol for "of" or "at" (as in $f$ of $x$) is very much needed in high school. Using blank space to mean an operation when it is the first time you are introducing that operation and it conflicts with another usage is just questionable.

The symbol $f(x)$ emerged rather late (in 1734 by Euler) and exclusively for the practicing mathematicians. The attempt to sell it to high school students appears as a bit of premature activity or oversight.

I have used something like Nick C. recalls.

$$ f\circ (\square) = 5\square^2+ 3\square +2 $$ In my precalculus and calculus I classes (instead of the standard $f(x)=5x^2+3x+2$). This is to emphasize two points, (a) $f$ is the name of a function, as flagged by $\circ$ and it takes an input, (b) the input is whatever is in $\square$. The box helps instruct the student to be careful with substitution. It also avoids the overuse of $x$.

We use blank space to indicate an operation in other places as well, for example when it comes to exponents $b^n$. This is not as much of a problem because it does not conflict with an earlier usage. However when it comes to $f^{-1}$ to mean functional inverse or $f^{(2)}$ to mean the second derivative we run into similar issues in calculus.

Answer 6

You start with $f(x)=x^2+x+1$. How about having students determine $f(y)$ where $y=z +1$?

First, substitute $y$ for $x$ to obtain $f(y)=y^2+ y +1$.

Then, substitute $z+1$ for $y$ to obtain $f(z+1)=(z+1)^2 +(z+1) +1$.

Then expand and collect like terms to obtain $f(z)=z^2 +3z +3$.

After a bit of practice, they learn that they can do two steps at once as a "shortcut."