Composition & Function Ops
This is your self-named biggest weakness — and it's one very specific habit, not a hole in your understanding. You know how to substitute. The trap is what you do after.
Composition vs. Operations
The order trap: $(f\circ g)(x)$ means "$g$ first, then $f$" — read it right to left, inside out. If you catch yourself plugging $f$ into $g$ instead, stop and reread which letter is on the outside.
One Question, 60 Seconds
Work this on paper first — full expansion, not a shortcut. Then check.
Find $(f \circ g)(x)$.
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B is correct. Substitute $(x+3)$ everywhere $f$ has an $x$: $(x+3)^2+1$. FOIL the binomial: $(x+3)(x+3)=x^2+6x+9$, plus 1.
If you picked A — this is the trap, named exactly: "square the front, square the back." $(x+3)^2$ is NOT $x^2+3^2$. A binomial squared has to be FOILed, every time. Go do the worked example below before the practice set.
If you picked C — you substituted but forgot to square the binomial at all.
If you picked D — you composed backwards: plugged $f(x)$ into $g$ instead of $g(x)$ into $f$. $g$ goes into $f$ — start from the inside.
Got B cold? Skim the worked example, then go straight to the practice set.
Worked Example 1 — Function Ops from a Graph
This part's already solid for you — one quick refresher on the order trap in division.
Reading a graph: $f(2) = 3$ and $g(2) = -1$. Find $(f/g)(2)$.
The trap: division isn't commutative — $3 \div (-1) \neq -1 \div 3$. Order matters. This is the same spot you slipped on a $2 \div -1$ problem before; the fix is just reading the operation left to right, same as it's written: $f$ first, then $g$.
Worked Example 2 — The Contrast That Matters
Same outer function, two different inner functions. Watch what changes.
Why these two side by side: the shortcut "square the front, square the back" happens to work when there's only one term inside $(3x)^2 = 9x^2$. The moment the inner function has two terms — $(x+4)^2$ — that shortcut breaks, and only FOIL gives the right answer. Same outer function, same-looking move, completely different result. Check which one you're looking at before you square anything.
Faded — Fill the Blanks
Find $(f\circ g)(x)$.
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$(x-5)^2-2 = (x^2-10x+25)-2 = x^2-10x+23$.
Practice Set (6) — Multiple Choice, Sheet Face-Down First
Work every problem in full on paper — expand it, don't shortcut it — before you look at the options.
$f(x)=x^2$, $g(x)=2x$. Find $(f\circ g)(x)$.
A) $2x^2$ B) $4x^2$ C) $x^2+2x$ D) $4x$
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B. $(2x)^2=4x^2$ — one term, no FOIL needed. This one confirms the substitution mechanic is solid before the trap shows up in #2.
$f(x)=x^2$, $g(x)=x+5$. Find $(f\circ g)(x)$.
A) $x^2+25$ B) $x^2+10x+25$ C) $x^2+5$ D) $2x+5$
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B. $(x+5)^2=(x+5)(x+5)=x^2+10x+25$. A is the "square the front, square the back" trap — same outer function as #1, but now there are two terms inside, so the shortcut breaks.
$f(x)=x^2+2$, $g(x)=x-1$. Find $(f\circ g)(x)$.
A) $x^2+3$ B) $x^2-2x+3$ C) $x^2-1+2$ D) $x^2-2x-1$
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B. $(x-1)^2+2 = (x^2-2x+1)+2 = x^2-2x+3$. Getting this one confirms the fix from #2 generalized — it wasn't just about that specific problem's numbers.
$f(x)=2x^2$, $g(x)=x+3$. Find $(f\circ g)(x)$.
A) $2x^2+9$ B) $2x^2+6x+9$ C) $2x^2+12x+18$ D) $2x^2+3$
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C. FOIL first — $(x+3)^2=x^2+6x+9$ — THEN multiply the whole thing by the leading coefficient 2: $2x^2+12x+18$. B is the trap of multiplying only the squared term by 2 and leaving the rest alone.
$f(x)=x^3$, $g(x)=x-2$. Find $(f\circ g)(x)$.
A) $x^3-8$ B) $x^3-6x^2+12x-8$ C) $x^3-2$ D) $x^3-2x$
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B. $(x-2)^3 = (x-2)(x-2)(x-2)$, fully expanded. "FOIL" only technically covers two binomials — the real rule is fully expand any power of a multi-term expression, not just "say FOIL and multiply two things." If A tempted you, that's the shortcut resurfacing in a new shape.
$f(x)=x^2-x$, $g(x)=x+1$. Find $(f\circ g)(x)$, then evaluate at $x=2$.
A) $x^2+x$, giving $6$ B) $x^2+x$, giving $8$ C) $x^2+x+2$, giving $8$ D) $x^2+x$, giving $4$
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A. $(x+1)^2-(x+1) = x^2+2x+1-x-1 = x^2+x$. At $x=2$: $4+2=6$. Cross-check it the other way, without using the composed formula at all: $g(2)=3$, then $f(3)=9-3=6$. Same number both ways — that's the self-check. Compose-then-evaluate and evaluate-then-compose have to agree; if they don't on test day, you made an error in one of the two paths, so redo whichever is faster to recheck.
Composition + Domain
A less common combination: finding the domain of a composed function. This wasn't the focus of your recent sessions, so treat this section as extra coverage, not a core requirement — but the mechanics are the same domain rules you already know, just applied twice.
$f(x) = \dfrac{1}{x}$, $g(x) = x-4$. Find the domain of $(f\circ g)(x)$.
A) $(-\infty,\infty)$ B) $(-\infty,4)\cup(4,\infty)$ C) $(-\infty,0)\cup(0,\infty)$ D) $x \neq -4$
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B. $g(x)=x-4$ has no domain restriction of its own, but $f$ needs its input $\neq 0$: $g(x) \neq 0 \Rightarrow x-4 \neq 0 \Rightarrow x \neq 4$. C mistakenly restricts $x$ instead of $g(x)$.
$f(x) = \sqrt{x}$, $g(x) = x-6$. Find the domain of $(f\circ g)(x)$.
A) $[0,\infty)$ B) $[6,\infty)$ C) $(-\infty,6]$ D) $(-\infty,\infty)$
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B. $f$ needs its input $\geq 0$: $g(x)\geq 0 \Rightarrow x-6\geq 0 \Rightarrow x\geq 6$. A applied the rule to $x$ directly instead of to $g(x)$ first.
Before You Leave — Add This to Your Sheet
Two lines: "Composition = plug all the way in, $g$ first (start from the inside), then FOIL if the inner function has more than one term — don't square the front and back separately." And: "Ops from graphs = read the $y$-values first, then do the arithmetic in the order it's written."