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Using Function Notation II


Alignments to Content Standards: F-IF.A.2

Task

Given a function $f$, is the statement $$f(x+h)=f(x)+f(h)$$ true for any two numbers $x$ and $h$? If so, prove it. If not, find a function for which the statement is true and a function for which the statement is false.

IM Commentary

The purpose of the task is to explicitly identify a common error made by many students, when they make use of the "identity" $f(x + h) = f(x) + f(h).$ A function $f$ cannot in general be distributed over a sum of inputs. This is an easy mistake to make because $$ f(x+h) = f(x) + f(h) $$ is a true statement if $f,x,h$ are real numbers and the operations implied by the parentheses are multiplication. The task has students find a single explicit example for which the identity is false, but it is worth emphasizing that in fact the identity fails for the vast majority of functions. Among continuous functions, the only functions satisfying the identity for all $x$ and $h$ are the functions $f(x)=ax$ for a constant $a$.

Solution

The statement is not true for all functions.

A function for which it holds is the function $f$ given by $f(a) = 5a$. If $f(a) = 5a$, then $$f(x+h) = 5(x+h) = 5 \cdot x + 5 \cdot h = f(x) + f(h).$$
A function for which the statement does not hold is the function $f$ given by $f(a) = a^2$. If $f(a) = a^2$, then $$f(x+h) = (x+h)^2=x^2 + 2xh + h^2.$$ This differs from $$f(x)+f(h)=x^2 + h^2$$ by $2xh$. This middle term is not zero unless $x$ or $h$ is zero.

Jason says:

over 5 years

The final sentence in the commentary would make more sense to me if the word "linear" were deleted from it.

Bill says:

over 5 years

Thanks Jason, I've deleted that word.

bworks says:

over 5 years

While this task seems very good, I believe the phase "given a function f" is ambiguous. The solution implies that f is a function of x, although the question was phrased "for any two numbers x and h." I believe the solution should highlight the difference between f(x+h) and f(x)+f(h) when both x and h are indeterminates, rather than having one the input of the function.

Michael Nakamaye says:

over 5 years

I think that there were two issues here, one of which was corrected and was a good catch! In the solution, the function f is now given as a function of a so that x and h are both treated equally and there is no possible confusion. Concerning f, it is a function or a map: it is not a function of any particular variable and in this case it was convenient to evaluate f at x+h where x and h are real numbers. I think and hope that this addresses these concerns.

Cam says:

over 5 years

Thanks, Mike. I think you isolated the issues nicely, and the fix looks good.

Cam says:

over 5 years

You bring up an important point. I agree that the wording should be tweaked to remove this amgbiguity, though I think it's beyond the scope of the problem to think of x and h as separate indeterminates, i.e., even implicitly thinking of the two-variable function g(x,h)=f(x+h). Let's see what others have to say.