Using Function Notation

Since Calculus makes heavy use of the idea of substituting something other than the variable into a function for purposes like evaluating certain important expressions like

f(x+h)-f(x)
h

which has some kind of silly purpose like finding a slope of a tangent line, most pre-calculus books have some exercises in this. For reasons that I still only partially understand, many students have a lot of trouble with this, so over the years I have come up with a few suggestions that seem to help.

Example 1

Problem: f(x)=x². Find f(x+1).

Solution: The idea here is to take simply take the function definition and replace each occurence of x with x+1, sort of like the Change All command in a word processor, but this doesn't quite work if you take it literally because then you get f(x+1)=x+1², which would mean that only the 1 was being squared rather than the whole x+1, so you either have to use a little common sense and mindfulness of the meaning or make another rule that on the right side of the equation you should always put parentheses around the x+1. After all unnecessary parentheses never hurt anything, and you can always remove them later when you see that they have no effect on the meaning. In fact even if you made the rule be that you replace the x's on both the right and left side with (x+1) you would end up all right. You would just have f((x+1))=(x+1)², and then you could see that one of the sets of parentheses could be removed without effecting the meaning.

But some students have trouble handling this for various reasons, some of which I partly understand and some I still don't at all. One important thing is to make sure to understand the basic idea of a function in the first place, that the f represents a rule or operation being applied its input and that the x is just a blank to be filled in. The f is not multiplying the x either, the f and the x are totally different kinds of creatures. The f is sort like a coke machine and the x is the 60¢--or whatever it's gone up to nowadays--and the x² is the can of coke.

Sometimes it helps to try to blur the x+1 in your mind to make sure you see it as a single thing. x+1 represents a number just as much as x does, so anything you can do with x you can do with x+1. It also helps to make a point of not trying to simplify at the same time you are substituting even if the expression you get is very ugly. Then after you have some kind of expression for your answer then simplify it. This way you won't have to think about function notation and algebra at the same time. An added advantage to this is that most instructors, including myself will probably give you at least a good amount partial credit for an unsimplified answer regardless how ugly it looks. Remember that the point of exercises like this is not to show your algebraic prowess, it is to show that you know how function notation works. When I give a problem like the one above, my main aim is not to see if you can expand out (x+1)²  by using FOIL or the squaring formula it is to see if you know that it represents (x+1)² rather than x²+1 or x²(x+1). I do normally require my students to simplify expressions that they get in these problems and many other instructor will too, because it is a good chance to keep in practice with your algebraic skills, but that is definitely the less important part.

Another thing that some students find that is helpful is to change the variable in the defining function, because it can be easier to let t be x+1 than to let x be x+1. So write f(t)=t² and say let t=x+1, what does this become. If that doesn't work try replacing the x with an empty pair of parentheses and writing f(      )=(      )². Then just fill in the parentheses with whatever you want. So if you want to complete f(x+1)=?, just take this definition and fill in the parentheses each with x+1 to get f(x+1)=(x+1)²

And yes, the final answer is x²+2x+1, for those of you who want to know whether you have your FOIL right.

Now for a harder problem just to show that f(x) isn't always x².

Example 2

Problem: f(x)=x³+5x²-6x+1. Find f(-x).

Solution: This one is a little trickier because there is more than one occurence of x, but the same rule holds. Just replace each occurence of x with -x and you are good to go. Then after that do any simplifying that you can do. If it helps, think of this as

f(    )=  (     )³+5(     )²-6(      )+1

Then fill in the parentheses with -x.

f(-x)=(-x)³+5(-x)²-6(-x)+1

Now simplify to get

f(-x)=-x³+5x²+6x+1.

Now here is an interesting one that might trick you if you are not careful, because it is too easy.

Example 3

Problem: f(x)=5. Find f(x³+5x²-6x+1).

Solution: f(x³+5x²-6x+1)=5
Reason: No blanks to fill in on the right side. f(      )=5 This is a constant function. No matter what you put into it you get 5. If you put a million into it you would get 5. If you put an elefant in it you would get 5, provided elefants were in its domain.