I thought I had a handle on function notation, but my brain keeps acting like the f is just something I can distribute or treat like a multiplier. Then I try a problem, and everything goes curly-brace-chaos in my notebook.
For example, if I’m told f(x) = 2x − 5 and asked for f(x+3), I keep doing this totally wrong thing: I write f(x+3) = 2x − 5 + 3 = 2x − 2. That feels so natural to me (like I’m just “adding 3 at the end”), but I’m pretty sure that’s not how it’s supposed to work. Why isn’t that valid? What exactly should I be replacing, and where?
Another spot I mess up is f(3x). My brain goes, “oh cool, that must be 3f(x).” So I do this very likely-wrong chain: f(3x) = 3f(x) = 3(2x − 5) = 6x − 15. Then I check the answer key and… welp, not matching. Where is that logic going off the rails? Is there a quick way to recognize when I’m allowed to pull a number out like that vs when I’m not?
I also trip over compositions. If f(x) = x^2 and g(x) = x + 1, I tried to do f(g(x)) and somehow ended up with x^2 + 1. I know that looks way too simple, but I don’t quite see what I’m missing in the “plug g into f” step. How should I be reading f(something) so I don’t lose track of parentheses and end up flattening everything?
Simple number example where I confuse myself: with f(x) = x^2, I wrote f(2) + 2 = f(4). That felt “balanced” in my head (add 2 outside vs inside), but I’m guessing that’s just me fooling myself. Along the same lines, I once claimed f(2x) = 2f(x). For x = 3, I compared f(6) and 2f(3) and confidently told myself they’re equal… which I’m now doubting.
Could someone explain, in a clear way, what f(x+h), f(ax+b), and f(g(x)) actually mean in terms of substitution? Like, step by step: what is being swapped in, and where do the parentheses go so I don’t start treating f like it’s “multiply by f”? Is there a small checklist or rule-of-thumb I can use to keep me from turning f(x+2) into f(x)+2? And how can I quickly sanity-check my result to catch these mistakes before I commit them in pen?
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
Think of f as a machine with x-shaped slots: f(•) means “take the whole inside expression and drop it into every x-slot, in parentheses, then simplify.” You never distribute f or treat it like multiplication. Example with f(x) = 2x − 5: f(x+3) = 2(x+3) − 5 = 2x + 6 − 5 = 2x + 1 (not 2x − 2), and f(3x) = 2(3x) − 5 = 6x − 5, which is not 3f(x) = 3(2x − 5) = 6x − 15. Composition is the same substitution idea: with f(x) = x^2 and g(x) = x + 1, f(g(x)) = f(x+1) = (x+1)^2 = x^2 + 2x + 1, while g(f(x)) = g(x^2) = x^2 + 1-two different machines in different orders. A quick worked example for f(ax+b): if f(x) = 2x − 5, then f(ax+b) = 2(ax+b) − 5 = 2a x + 2b − 5; just plug ax+b wherever x appears. Sanity checks save lives: with f(x) = x^2, f(2)+2 = 4+2 = 6 but f(4) = 16, so f(x)+2 ≠ f(x+2); and f(2x) = (2x)^2 = 4x^2, while 2f(x) = 2x^2-only equal when x=0. Rule of thumb: to compute f(stuff), copy the formula for f, wrap each x in parentheses, and replace each x with (stuff), then simplify; if you’re tempted to “pull a number out,” stop and test with a quick value of x (like x=1 or x=2). The only times “pull out” tricks work universally are for special linear forms: f(x+a) = f(x)+a holds only when f(x) = x + c, and f(ax) = a f(x) holds only when f(x) = m x (no constant term). Keep the parentheses discipline, and you’ll tame the curly-brace chaos!
Rule of thumb: read f(□) as “take the recipe for f and pour □ into every x‑slot (with parentheses!)”-f isn’t a multiplier you can distribute.
Example buffet: with f(x)=2x−5, we get f(x+3)=2(x+3)−5=2x+1 (not f(x)+3), f(3x)=2(3x)−5=6x−5 (not 3f(x)); with f(x)=x^2 and g(x)=x+1, f(g(x))=(x+1)^2; quick check: f(2)+2=4+2=6 while f(4)=16, so “inside vs outside” aren’t equal-test a number (say x=3) to catch slips.
Think of f as a rule that eats whatever is inside the parentheses. To compute f(something), take the defining formula for f and replace every x with that entire “something,” keeping parentheses around it. So with f(x) = 2x − 5: f(x+3) = 2(x+3) − 5 = 2x + 1, and f(3x) = 2(3x) − 5 = 6x − 5. That’s why f(3x) is not 3f(x) here: 3f(x) = 3(2x − 5) = 6x − 15. Those only match if the function has no constant term (e.g., f(x) = ax). A quick self-check is to test a simple x, like x = 0; if you think f(x+3) = f(x) + 3, then LHS = f(3) = 1 while RHS = f(0) + 3 = −2, so the claim fails immediately.
For composition, f(g(x)) means plug the whole g(x) into f. If f(x) = x^2 and g(x) = x + 1, then f(g(x)) = [g(x)]^2 = (x+1)^2 = x^2 + 2x + 1 (the middle term appears because of the square). The same “inside vs outside” idea explains common slips: with f(x) = x^2, f(x+2) = (x+2)^2 = x^2 + 4x + 2^2, whereas f(x) + 2 = x^2 + 2; and f(2x) = (2x)^2 = 4x^2, whereas 2f(x) = 2x^2. Numbers inside the parentheses change the input before the rule is applied; numbers outside change the output after.
A simple checklist I use: first, rewrite f as a template with a blank, like f(□) = 2□ − 5 or f(□) = □^2. Next, drop the entire new input into that blank with parentheses, then simplify. Finally, sanity-check by trying an easy value (x = 0 or 1) to catch “pulling numbers out” mistakes. Does the “template” trick make the substitution step feel clearer? If so, want to try it on f(x) = 3x^2 − x + 4: what are f(x+1), f(2x), and f(g(x)) when g(x) = 1 − x?