In calculus, we will need to take a function *f *(*x*) and write out *f *(*x*+*h*) for that function. Let’s look at how to do this properly.

To do a problem like this, you need to understand exactly what the *x* in *f *(*x*) represents and what the *f* represents. Let’s look at the function *f *(*x*) = *x*^{2} – *x.* A function is a process. In this case, it is the process of

- Square the input
- Take the result and subtract the input

Notice that there is no mention of the *x* in the formula. That is because it is a placeholder representing the input. There is nothing special about x. We could have just as easily used a different letter as a placeholder for the input. If I had wanted to call the input *t*, I would have written

*f *(*t*) = *t*^{2} – *t*

If the input had been represented by the word *dog*, I would have written

*f *(*dog*) = *dog*^{2} – *dog*

The input variable is simply a placeholder…if a number is put in its place like 7, we get

*f *(*7*) = 7^{2} – 7* = 42*

Notice that the process is the same. Square the input and subtract the input from the result. In this case, the input is 7 so we are squaring 7 and then subtracting 7 from the result.

Many students are confused by *f*(x+*h*). Now the input is represented by *x*+*h* instead of *x*. This means we need to square it and then subtract x+h from the result.

*f *(*x+h*) = (*x+h)*^{2} – (*x+h)*

We can simplify this by foiling out the square,

(*x*+*h*) = (*x*+*h*)(*x*+*h*) = *x*^{2} +2*xh* + *h*^{2}

And removing the parentheses after the subtraction we get

*f *(*x*+*h*) = *x*^{2} +2*xh* + *h*^{2} – *x* – *h*

The handout below has more examples with this function.