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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?

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    $\begingroup$ How are they with $f(2x)$? $\endgroup$
    – TomKern
    Feb 15, 2023 at 21:13
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    $\begingroup$ @TomKern I think a good answer to the question could suggest a better sequence of diagnostics like the one you suggest. I usually go from $f(3)$ to $f(17)$ to $f(☺)$ to $f(\text{Tomato Soup})$ to $f(y)$ to $f(x+h)$. But it's possible I could gain critical insight from some better sequence of diagnostic questions. $\endgroup$ Feb 16, 2023 at 0:05
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    $\begingroup$ Perhaps the problem is that not everyone sees that what is inside the brackets of the function is a single expression, that must be evaluated before the function is applied? That is, students might not see that $f(x+h)$ actually means $f\bigl(\thinspace(x+h)\thinspace\bigr)$, i.e. given $x$ and $h$ you must first add them to obtain "$x+h$" and then apply $f$ to the result of that? $\endgroup$
    – printf
    Feb 16, 2023 at 5:40
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    $\begingroup$ One question that comes to mind is whether your oral presentation is helping or hurting here. In particular, when talking to your students (casually, without trying to emphasize the difference), do you pronounce $f(x+h)$ and $f(x)+h$ the same way or differently? $\endgroup$ Feb 16, 2023 at 14:21
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    $\begingroup$ I'm trying to be polite about this but can you all just post answers instead of answers in the comments? $\endgroup$ Feb 16, 2023 at 21:57

6 Answers 6

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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...

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    $\begingroup$ Ooh -- I like these a lot. I was definitely lacking diagnostics between "xxx" and "x+h." I am definitely fine with improving my diagnostics even if the cure rate is low, too. :) $\endgroup$ Feb 16, 2023 at 0:00
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    $\begingroup$ A warning: if you apply these diagnostics in order you will have a lot more students who pass than if you apply one at random. The "scaffolding effect" can get them there without any real understanding. $\endgroup$ Feb 16, 2023 at 1:33
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    $\begingroup$ @StevenGubkin I suppose if the goal is to teach them, that's fine; if the goal is to understand the teaching and learning process, it's not $\endgroup$
    – user253751
    Feb 16, 2023 at 14:31
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    $\begingroup$ I think this kind of scaffolding can be an effective teaching technique, but you need to gradually remove the "supports". It can be tempting, as an instructor, to see your students' success after scaffolding and trick yourself into thinking that they now understand. $\endgroup$ Feb 16, 2023 at 14:37
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    $\begingroup$ Great answer and great question. Is there any place that actually conceptually thinks in this way about math education? I mean the diagnostic approach? It seems really smart. $\endgroup$
    – DRF
    Feb 18, 2023 at 9:53
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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.

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    $\begingroup$ This is nice, at least for functions of one variable! $\endgroup$ Feb 15, 2023 at 21:56
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    $\begingroup$ This is interesting. Of course to me, $f(x) = 5x^2 - 3x + 1$ looks very similar to $f(---) = 5(---)^2 - 3(---) + 1$, because I see the $x$ as a "blank" already. Perhaps many students have specialized "$x$" and do not see it as a blank. It seems like this concept should have a name... $\endgroup$ Feb 16, 2023 at 0:04
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    $\begingroup$ @ChrisCunningham In my personal experience, the concept of x being a "blank" is conveyed very poorly in pre-calculus classes. It is explained like once when the concept of functions is introduced, and then they use x in every single expression ever as the default sign of function variable. With this background, for many students it is quite nontrivial to realize that the x in f(x) can actually be anything, a variable can be denoted with something other than x, and that a variable denoted with x is not special at all unless told so. $\endgroup$
    – Neinstein
    Feb 16, 2023 at 14:53
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    $\begingroup$ I had an instructor that made us use ( ) this way. You wouldn't see underlines. You had to watch it being written out to see it like this because when you're done it just looks normal. But to this day I think of parenthesis as things you drop "inputs" into. $\endgroup$ Feb 16, 2023 at 18:16
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    $\begingroup$ @ChrisCunningham I think you have a key gap in your model of where students are coming from which this answer(+1) indirectly addresses, but which I'd like to call out explicitly: Students who can evaluate f(🙂) understand the x as a blank they can replace with anything, but all of the tests in your question are with expressions that basically don't require parentheses. And similarly most evaluation of functions in their prior experience is with numbers or single letters or maybe a power, none of which require them to invent or practice introducing parentheses in their evaluation of functions. $\endgroup$
    – Mark S.
    Feb 17, 2023 at 8:59
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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}

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    $\begingroup$ This answer seems to me like a very good hypothesis for why $f(☺)$ is easy but $f(x+h)$ is difficult. To test this more precisely, I should try asking students to find $f(y+h)$, right? $\endgroup$ Feb 16, 2023 at 18:22
  • $\begingroup$ @ChrisCunningham Yes, that might be a good way to test it. $\endgroup$
    – user76284
    Feb 16, 2023 at 18:27
  • $\begingroup$ Excellent answer! $\endgroup$ Feb 16, 2023 at 21:21
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    $\begingroup$ I think @Steven's answer is awesome, but yours is even more satisfying because it appears to be stating the obvious... so obvious that it ironically resides in our blind spots. $\endgroup$
    – ryang
    Feb 17, 2023 at 6:51
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    $\begingroup$ Come to think of it, how heretical actually is it to write sentences like $g(x)=\displaystyle\int_0^x f(x) \textrm{ d}x$ and $h(k)=k\displaystyle\sum_{k=3}^7 a_k$ ? They're certainly shoddy, but is $g(5)=\int_0^5 f(x) \textrm{ d}x$ or $h(5)=5\displaystyle\sum_{k=3}^7 a_k$ more wrong than $\forall xPx\land\exists xQx\equiv \forall xPx\land\exists yQy$ ? @StevenGubkin $\endgroup$
    – ryang
    Feb 17, 2023 at 12:39
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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.

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    $\begingroup$ I like the scribble idea. The strange thing is that the $x$ is supposed to be that scribble already! So it seems like a symptom of something very strange that we somehow wore out $x$, giving it too much meaning, when it was supposed to be a scribble all along... :) $\endgroup$ Feb 16, 2023 at 16:07
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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.

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    $\begingroup$ The question says they handled Tomato Soup, $xxx$, and $y$ correctly, so the problem doesn't seem to be misinterpreting function application. The problem seems to be substitution (see my answer). $\endgroup$
    – user76284
    Feb 16, 2023 at 18:14
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    $\begingroup$ This is an insightful answer and I'm glad to see it, but it does not line up well with the evidence I have collected. Specifically, it is very rare for a student to see $f(x) = x^2 + x + 1$ and then compute $f(x+h) = (x^2+x+1)(x+h)$, mistaking the function parentheses for multiplication. It's very possible this causes problems elsewhere in the curriculum though. $\endgroup$ Feb 16, 2023 at 18:24
  • $\begingroup$ @user76284 In$ f(x+h)$ student has to process two functions, the first is the instruction to add $x$ and $h$, the second is to apply $f$ to the output. This complication is not seen when the input is Pizza or xxx. There is nothing to process there. So there is less change of confusion. (If a recipe for an actual pizza is given the student, and I for that matter, are likely to be confused with the order of operation there as well!) $\endgroup$
    – Maesumi
    Feb 17, 2023 at 18:05
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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."

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