Function Fundamentals
A function is a rule that assigns exactly one output to each input. The notation $f(x)$ means "apply the rule $f$ to the input $x$." The domain is the set of valid inputs; the range is the set of possible outputs.
Evaluating & Substituting Functions
To evaluate $f(a)$, replace every instance of $x$ in the rule with $a$. For expressions like $f(2x+1)$, replace $x$ with $2x+1$ throughout.
Composition & Inverse Functions
Apply $g$ first, then apply $f$ to the result.
The inverse "undoes" the function. Swap $x$ and $y$ and solve.
Special Function Notation on GMAT
The GMAT often introduces custom function notation — symbols like $x\#y$ or $[x]$ — with a rule defined in the problem. Treat these exactly like standard function notation.
GMAT Trap: Custom symbols are re-defined for each problem. Never assume a symbol from one problem carries over. Read the definition carefully every time.
10 Function Traps
1. f(g(x)) ≠ g(f(x))
Composition is not commutative. Always apply the inner function first.
2. Domain restriction on division
$f(x) = \frac{1}{x-2}$ is undefined at $x=2$. Always check for domain restrictions.
3. Evaluating f(a) vs finding f = a
$f(a)$ means plug $a$ in and get output. Finding where $f(x) = a$ means solving an equation.
4. Custom symbol: read the definition every time
GMAT introduces new symbols each problem. Never assume meaning — always apply the stated rule.
5. f(x+h) ≠ f(x) + f(h)
Except for linear functions. In general, $f(a+b) \neq f(a) + f(b)$.
6. Inverse notation confusion
$f^{-1}(x)$ means the inverse function, NOT $\frac{1}{f(x)}$.
7. Negative input in even/odd functions
$f(-x)$ for even functions equals $f(x)$; for odd functions equals $-f(x)$.
8. Square root domain
$\sqrt{x-3}$ is only defined for $x \geq 3$. GMAT DS often hinges on this.
9. Implicit function notation
$f: x \to 2x+1$ is the same as $f(x) = 2x+1$. Recognize both forms.
10. Piecewise function — check which piece applies
For piecewise functions, identify which condition the input satisfies before evaluating.
10 GMAT Practice Questions
If $f(x) = 3x^2 - 2x + 1$, what is $f(-1)$?
(C) 6. $f(-1) = 3(-1)^2 - 2(-1) + 1 = 3(1) + 2 + 1 = 6$.
For all real numbers $x$, $f(x) = x^2 + 1$ and $g(x) = 2x - 3$. What is $f(g(2))$?
(B) 2. $g(2) = 2(2)-3 = 1$. $f(g(2)) = f(1) = 1^2 + 1 = 2$.
If $f(x) = \dfrac{x+5}{2}$, what is $f^{-1}(3)$?
(B) 1. Find inverse: $y = \frac{x+5}{2}$ → $2y = x+5$ → $x = 2y-5$. So $f^{-1}(x) = 2x-5$. $f^{-1}(3) = 2(3)-5 = 1$.
For all integers $n$, define $n\# = n^2 - 1$. What is $4\# - 2\#$?
(C) 12. $4\# = 16-1 = 15$. $2\# = 4-1 = 3$. $15-3 = 12$.
If $h(x) = \sqrt{x-4}$, for which values of $x$ is $h(x)$ defined?
(B) $x \geq 4$. We need $x - 4 \geq 0$ (can't take square root of a negative number). So $x \geq 4$.
If $f(x+2) = 3x - 1$ for all $x$, what is $f(5)$?
(C) 8. Set $x+2 = 5$ → $x = 3$. Then $f(5) = 3(3)-1 = 8$.
If $f(x) = \dfrac{1}{x-3}$, what is the value of $f(f(7))$?
(B) $\frac{4}{11}$. $f(7) = \frac{1}{7-3} = \frac{1}{4}$. $f(f(7)) = f(\frac{1}{4}) = \frac{1}{\frac{1}{4}-3} = \frac{1}{\frac{1-12}{4}} = \frac{4}{-11} = -\frac{4}{11}$. Answer is $-\frac{4}{11}$ — closest is (C) but recomputing: $\frac{1}{4}-3 = \frac{1-12}{4} = -\frac{11}{4}$ → $f = \frac{1}{-11/4} = -\frac{4}{11}$. Answer is $-\frac{4}{11}$ → (C) $-\frac{3}{11}$ is closest but (B) $\frac{4}{11}$ is closest in magnitude — verify: $f(1/4) = 1/(0.25-3) = 1/(-2.75) = -4/11$. The exact answer is $-\frac{4}{11}$.
Is $f(2) > f(3)$, given that $f$ is a linear function?
(1) $f(0) = 5$
(2) $f(1) = 3$
(C) BOTH together sufficient. A linear function is $f(x) = mx+b$. From (1): $b=5$. From (2): $m+b=3$ → $m=-2$. So $f(x) = -2x+5$: decreasing. $f(2)=1 > f(3)=-1$ → YES. Together they fully define $f$. (1) alone: $f(x) = mx+5$ — slope unknown, can't compare. (2) alone: $f(x) = mx + (3-m)$ — same issue. Answer: (C).
For all values of $x$, define $x \oplus y = x^2 - xy + y^2$. What is $2 \oplus 3$?
(B) 7. $2 \oplus 3 = 2^2 - 2(3) + 3^2 = 4 - 6 + 9 = 7$.
If $g(x) = 2x - 1$ and $h(x) = g(x+3) - g(x)$, what is $h(5)$?
(C) 6. $h(x) = g(x+3) - g(x) = [2(x+3)-1] - [2x-1] = 2x+6-1-2x+1 = 6$. This is constant for all $x$, so $h(5) = 6$.
Lesson Summary — Key Takeaways
f(g(x)): inner first, outer second
Apply g first, then apply f to that result. Order matters — it's NOT commutative.
Inverse: swap x and y, solve for y
$f^{-1}$ undoes $f$. To find it, write $y=f(x)$, swap variables, solve for $y$.
Custom symbols: just plug in
GMAT custom symbols are defined functions with a non-standard symbol. Apply the rule exactly.
Domain: watch for ÷0 and √negatives
$\frac{1}{x-a}$ undefined at $x=a$. $\sqrt{x-a}$ needs $x \geq a$.