I’m revising my algebra fundamentals, and function notation keeps bonking me on the nose like a curious cat. I get that f(x) is like a little machine, but when I try to feed it different snacks, I’m not sure what I’m actually giving it.
Could someone explain, in plain terms, the difference between plugging into a function and multiplying the function? For example, if f(x) = 2x + 5:
– What’s the real difference between f(3), f(3x), and 3f(x)? My brain keeps insisting f(3x) should be the same as 3f(x), but I have a sneaky feeling that’s the mathematical equivalent of mistaking a toaster for a fax machine.
Similarly, with the same f(x):
– Is f(x+2) generally the same as f(x) + 2? If not, how can I tell quickly? For a concrete spot-check, I tried looking at x = 4 and comparing f(x+2) to f(x) + 2, but I’m not sure if I’m thinking about it the right way.
And then composition vs multiplication:
– If g(x) = x − 1 and f(x) = x^2, what exactly is happening in f(g(x)) compared to f(x)g(x)? I tried to ‘distribute’ f over things (like f acting as a friendly octopus putting arms on sums and constants), but I don’t think functions distribute like that. Is there a simple rule-of-thumb to stop me from doing this?
I’m trying to strengthen my basics, so I’d love a clear way to read these notations and know what operation I’m actually doing. I tried expanding a few expressions, but I’m not sure if that was even relevant to the misunderstanding. How should I think about f(3), f(3x), 3f(x), f(x+2) vs f(x)+2, and f(g(x)) vs f(x)g(x) without falling into the ‘distribute the f’ trap?
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3 Responses
Think of f as a machine with one input slot: whatever you put inside the parentheses replaces x everywhere in its recipe. So “plugging” means substitution, while multiplying the function means scaling its output after it finishes. For f(x)=2x+5, we get f(3)=2·3+5=11 (a number out), f(3x)=2·(3x)+5=6x+5 (a new function in x), and 3f(x)=3·(2x+5)=6x+15 (also a new function), so f(3x) ≠ 3f(x) in general; they only match for special “through-the-origin” linear cases like f(x)=ax. Similarly, inside vs outside matters: f(x+2)=2(x+2)+5=2x+9 shifts the input before the machine works (a horizontal effect), while f(x)+2=(2x+5)+2=2x+7 lifts the output after the fact (a vertical effect). These are equal for all x only when f(x)=x+C. For composition, with g(x)=x−1 and f(x)=x², f(g(x)) means “run g first, then feed its output to f,” giving (x−1)², while the product f(x)g(x) multiplies the two separate outputs, giving x²(x−1); very different operations. A handy rule-of-thumb: never “distribute the f” over sums or scalars-generally f(a+b) ≠ f(a)+f(b) and f(cx) ≠ c·f(x)-unless you know f is linear in the strict sense with f(0)=0. When in doubt, do a quick substitution test (try x=0 or x=1): if the results differ even once, the expressions aren’t the same. Analogy: imagine a smoothie blender-changing what you toss in (x → 3x or x+2) changes the flavor before blending, while multiplying the final smoothie by 3 just means you poured three identical servings after it was blended.
Oh, I love this question. Function notation can feel like a mischievous cat that keeps batting at the parentheses. Let’s tame it.
Golden rule (tattoo this on your math brain): f( … ) means “take the recipe for f and wherever you see x, plug in the entire thing inside the parentheses.” After that, anything multiplying f(x) on the outside is scaling the output, not changing the input.
Let’s use your examples one by one, with f(x) = 2x + 5.
1) f(3), f(3x), and 3f(x)
– f(3): Plug in x = 3.
f(3) = 2·3 + 5 = 11. That’s just a number.
– f(3x): Plug in the whole expression 3x for x.
f(3x) = 2·(3x) + 5 = 6x + 5. That’s an expression in x.
– 3f(x): First compute f(x), then multiply that result by 3.
3f(x) = 3·(2x + 5) = 6x + 15.
Notice the difference:
– f(3x) = 6x + 5
– 3f(x) = 6x + 15
They look similar but are not the same. The +5 got multiplied by 3 only in 3f(x), because 3 is outside the function (scaling the output). Inside the parentheses changes the input; outside multiplies the output.
When would f(3x) = 3f(x) hold? For a function that’s “truly linear” in the linear-algebra sense: f(x) = ax with no constant term. If f(x) = ax + b, then
– f(3x) = 3ax + b
– 3f(x) = 3ax + 3b
These are equal for all x only if b = 0.
2) f(x+2) vs f(x) + 2
Let’s compute both with f(x) = 2x + 5.
– f(x+2): Plug in x+2.
f(x+2) = 2(x+2) + 5 = 2x + 9
– f(x) + 2: Compute f(x) and then add 2.
f(x) + 2 = (2x + 5) + 2 = 2x + 7
So, for this f, f(x+2) ≠ f(x) + 2.
Quick spot-check method: try a single x. If you get different values, they’re not equal as functions. For example, x = 4:
– f(x+2) = f(6) = 2·6 + 5 = 17
– f(x) + 2 = f(4) + 2 = (2·4 + 5) + 2 = 13 + 2 = 15
Since 17 ≠ 15, they’re different.
When could f(x+2) = f(x) + 2 be true for all x? If f(x) = ax + b, then
– f(x+2) = a(x+2) + b = ax + 2a + b
– f(x) + 2 = ax + b + 2
These are equal for all x only if 2a + b = b + 2, i.e. a = 1. So f(x) = x + b works; your f(x) = 2x + 5 does not.
3) Composition vs multiplication: f(g(x)) vs f(x)g(x)
Let g(x) = x − 1 and f(x) = x^2.
– f(g(x)): Do g first, then feed that into f.
g(x) = x − 1, so f(g(x)) = (x − 1)^2 = x^2 − 2x + 1.
– f(x)g(x): Multiply the two outputs.
f(x)g(x) = (x^2)(x − 1) = x^3 − x^2.
Totally different creatures. Composition is nesting: replace x in f by g(x). Multiplication is just a product of two numbers/expressions after both functions are evaluated at the same x.
If you’re curious (tiny tangent): g(f(x)) is different again:
g(f(x)) = g(x^2) = x^2 − 1.
A small rule-of-thumb to avoid the “distribute the f” trap
– You can only “distribute” in the sense f(u+v) = f(u) + f(v) and f(cu) = c f(u) if f is a linear map (in the strict sense) with f(x) = ax and f(0) = 0. Most everyday functions (x^2, 2x+5, sin x, etc.) are not linear in that sense, so no distributing.
– Read parentheses as “this is the input.” What’s inside f( … ) changes the input. Constants or multipliers outside f( … ) change the output.
A simple worked example cheat-sheet with f(x) = 2x + 5
– f(3) = 11
– f(3x) = 6x + 5
– 3f(x) = 6x + 15
– f(x+2) = 2x + 9
– f(x) + 2 = 2x + 7
– With g(x) = x − 1 and f(x) = x^2:
– f(g(x)) = (x − 1)^2 = x^2 − 2x + 1
– f(x)g(x) = x^2(x − 1) = x^3 − x^2
Fast test habits
– To see if two expressions define the same function, you can try a quick plug-in like x = 0 or x = 1; if you get different outputs, they’re different. (One counterexample is enough.)
– To prove they’re the same for all x, do the substitution algebraically (replace x by the whole input) and simplify until the two sides match.
If you keep the mantra “inside the parentheses changes the input; outside changes the output,” the cat usually stops bonking your nose. And if it starts again, toss it a quick test value like x = 0 and see what happens.
Think of f as a rule you apply to an input. A handy habit: write f with a blank box and drop the whole input in. For f(x) = 2x + 5, that’s f(□) = 2□ + 5. Then:
– f(3) = 2·3 + 5 = 11.
– f(3x) = 2·(3x) + 5 = 6x + 5.
– 3f(x) = 3·(2x + 5) = 6x + 15.
So f(3x) and 3f(x) are not the same (the +5 gets scaled in 3f(x) but not in f(3x)). Similarly, f(x+2) = 2(x+2) + 5 = 2x + 9, while f(x) + 2 = (2x + 5) + 2 = 2x + 7; they differ. Quick spot-check at x = 4: f(6) = 17, but f(4) + 2 = 13 + 2 = 15. Rule of thumb: f(x+2) is a horizontal shift of the input (left by 2), while f(x)+2 is a vertical shift of the output (up by 2). They only match for very special f (e.g., f(x) = x + c).
For composition vs product, “compose” means substitute, “product” means multiply outputs. With g(x) = x − 1 and f(x) = x^2, we have f(g(x)) = (g(x))^2 = (x − 1)^2 = x^2 − 2x + 1, while f(x)g(x) = (x^2)(x − 1) = x^3 − x^2. Very different. Rule of thumb to avoid the “distribute the f” trap: never split f across sums or constants; just replace every x in the formula for f with the entire inside expression once. In symbols: f(□) = 2□ + 5 or f(□) = (□)^2, then drop the inside in for □. Distributive-looking identities like f(a + b) = f(a) + f(b) or f(cx) = c f(x) only hold for special linear functions; don’t assume them by default.