f(x) names the output that the function f produces from the input x. It is one number, not f multiplied by x, and once that is settled the notation distinguishes two different questions: evaluate a given input, and solve for the inputs that produce a given output.
Core ideas
“What a Function Is” established that an input has exactly one output. This lesson gives that output a name, which is what makes every later lesson writable.
f(3) is a single value. The letter f is the name of the rule, the brackets are not multiplication, and x is a slot. Read f(3) as "f of 3", never as "f times 3", because the second reading makes f(a + b) = fa + fb look like the distributive law.
To find f(−2) for f(x) = x² − 4x + 5, write (−2)² − 4(−2) + 5. Every occurrence of x becomes (−2), brackets and all. The brackets are what stop (−2)² from collapsing to −4.
f(3) hands you an input and asks for the output; there is exactly one answer. g(x) = 5 for g(x) = 3x − 7 hands you an output and asks which input produced it; Linear Equations and Inequalities's linear method gives x = 4. On a table or graph there may be none, one or several such inputs. Confusing the two questions is the most common error in this lesson.
The material
What the symbol means, how to substitute safely, and the two questions it can pose.
If f is a function and x is an input in its domain, f(x) denotes the output that f assigns to x. The letter naming the function and the letter naming the input are both arbitrary: f(x) = 2x + 1, g(t) = 2t + 1 and h(u) = 2u + 1 are three names for one function.
The brackets are borrowed from the brackets of multiplication, and that is an accident of history, not a hint. Nothing in f(x) is being multiplied. The notation is roughly three hundred years old and predates any convention that would have made it unambiguous, so the ambiguity is permanent and the only defence is reading the whole symbol as one name.
Take f(x) = x² − 4x + 5. Substituting means replacing every x with the input, wrapped in brackets, and only then simplifying.
The third line is where brackets earn their place. Without them the first term reads −2², which is −4, and the second reads −4·2, which is −8; the total comes out −7 instead of 17. Both slips are silent, because −7 is a perfectly ordinary number and nothing in the working looks wrong.
The slot accepts anything, including another expression. The procedure does not change: replace every x by the whole expression in brackets, expand, collect.
Compare that with f(a) + f(1), which is a² − 4a + 5 plus 2, or a² − 4a + 7. The two differ at every value of a except where a² − 2a + 2 = a² − 4a + 7, that is 2a = 5, so they agree only at a = 5/2. A function does not distribute over addition, and a single accidental agreement is not evidence that it does.
For a general function f, f(a + b) and f(a) + f(b) are different numbers. The error survives because it is correct for f(x) = cx, the family met first: c(a + b) = ca + cb really is the distributive law. A constant term is enough to break it. For f(x) = 2x + 1, f(1 + 1) = 5 while f(1) + f(1) = 6.
A point (3, 2) on the graph of f says f(3) = 2, because the graph is the set of pairs (input, output). Finding f(3) means reading up from 3 on the horizontal axis; solving f(x) = 2 means reading across from 2 on the vertical axis and collecting every x beneath. One vertical sweep, one horizontal sweep, and they can return different numbers of answers.
Interactive
One input in, one output out, substituted in full
Method
Four moves. The third is the one that goes wrong.
An input inside the brackets means evaluate. An equals sign after the brackets means solve. Answering the wrong one produces a confident, irrelevant number.
For f(x) = x² − 4x + 5, write ( )² − 4( ) + 5. This takes five seconds and makes it impossible to substitute into some occurrences of x and not others.
(−2)², not −2². −4(−2), not −4−2. When the input is negative or is itself an expression, the brackets are carrying the arithmetic, not decorating it.
For a symbolic answer such as f(a + 1) = a² − 2a + 2, put a = 2 in both the original and the result. If f(3) and the new expression disagree, the expansion is wrong and you have found it in one line.
Use f(x) = x² − 4x + 5 for substitution and g(x) = 3x − 7 for a solve that needs only Linear Equations and Inequalities.
Questions 1 and 3 are the pair to keep apart. Both mention f, x and a number; one starts with an input and returns its output, while the other starts with an output and recovers its input.
Question 4 is worth comparing with f(a) + f(1) = a² − 4a + 7. The two expressions agree only at a = 5/2, so a spot check at that one value would wrongly suggest the shortcut works — which is why the check above uses a = 2.
For a faded pass, let p(t) = t² + 3t − 1. The bracketed substitution for p(−2) is supplied as (−2)² + 3(−2) − 1; simplify it and check the sign of each term. Then form p(a − 1) by replacing every slot before expanding. Separately solve 4x + 3 = 19 as the function equation q(x) = 19 for q(x) = 4x + 3.
Given r(x) = x² + 2x − 3 you should evaluate r(−4) and form r(a + 2) from bracketed substitutions, then solve the separate linear function equation s(x) = 11 for s(x) = 2x − 5. For each answer, say whether it is an output or an input and check it in the original rule.
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