GMAT Focus Edition · Advanced Quantitative Aptitude
Functions, Graphs and Maxima-Minima
Function types, composition, graphs, modulus, and optimization problems.
Eight concepts. CAT algebra here is less about memorising definitions and more about reading a graph, composing in the right order, and spotting where a maximum hides.
- GMAT Focus Edition
- Hard level
- 8 concepts
- 21 practice questions
1What makes a function
A relation is a function when every input in the domain maps to exactly one output. The vertical-line test is the picture of that rule: if any vertical line meets the graph more than once, two outputs share an input and the relation fails. A function is injective (one-to-one) when distinct inputs never share an output — equivalently, every horizontal line meets the graph at most once.
Figure. A vertical line meets y=x^2 once — one output per input — so that curve is a function. The same line would cut the dashed sideways parabola x=y^2 twice, so that relation fails the vertical-line test.
The set of pairs \{(1,2),(1,3),(4,5)\} describes
- A function that is not injective
- A relation that is not a function
- An injective function
The input 1 appears twice with different outputs, so the relation fails the function rule before injectivity is even asked.
2Composition
Composition stacks functions inside-out: (f\circ g)(x)=f(g(x)), so g runs first and its output feeds f. Order matters — in general the two compositions differ, and CAT questions often trap you by swapping the order while keeping the same symbols on the page.
Figure. (f\circ g)(x)=f(g(x)) runs g first (inner), then feeds the result into f (outer). Swapping the boxes is a different function.
Evaluating a composition
- Inner firstEvaluate g at the given input and simplify.
- Outer secondSubstitute that value into f.
- Check orderIf the question asked for g\circ f, the inner function would be f instead — reread the circle notation before calculating.
Evaluating a composition
If f(x)=2x+3 and g(x)=x^2, find (f\circ g)(2).
- g(2) = 2^24
- f(4) = 2\cdot4 + 311
- (f\circ g)(2)11
Pro tip. Always work inside-out. A common slip is computing f(2) first and then squaring — that is g\circ f, not f\circ g.
With the same f(x)=2x+3 and g(x)=x^2, the value of (g\circ f)(2) is
- 11
- 49
- 7
Here f is inner: f(2)=7, then g(7)=49. The composition example above computed (f\circ g)(2)=11 — same functions, reversed order.
3Inverse functions
An inverse exists only when the mapping is a bijection — both a function and injective — so it can be reversed without ambiguity. Write it as f^{-1}; the defining identity is f(f^{-1}(x))=x and f^{-1}(f(x))=x on the relevant domains. On a graph, f and f^{-1} are mirror images across the line y=x.
Figure. The solid curves are reflections across the dashed line y=x. The point (1,1) lies on f, on f^{-1}, and on the mirror line — that is what f(f^{-1}(1))=1 looks like.
Finding an inverse
- Write y=f(x)Replace f(x) by y.
- Swap and solveExchange x and y, then solve for y.
- VerifyCompose with the original — both f(f^{-1}(x)) and f^{-1}(f(x)) should collapse to x.
Inverse of a linear function
Find f^{-1}(x) if f(x)=3x-2, and verify f(f^{-1}(7))=7.
- y = 3x - 2 \Rightarrow x = 3y - 2swap variables
- 3y = x + 2y = \frac{x+2}{3}
- f\bigl(f^{-1}(7)\bigr) = 3\cdot\frac{7+2}{3} - 27
Pro tip. If solving for the inverse gives multiple branches, the original was not one-to-one on the stated domain — restrict the domain before inverting.
4Even and odd symmetry
Even functions satisfy f(-x)=f(x) and their graphs are symmetric about the y-axis — think x^2 or \cos x. Odd functions satisfy f(-x)=-f(x) and are symmetric about the origin — think x^3 or \sin x. Most functions are neither; do not assume symmetry without checking both signs.
Figure. The even curve folds like a mirror across the y-axis; the odd curve rotates 180^\circ about the origin. Both pass through (0,0), but only the odd one changes sign when x does.
Testing symmetry
Classify f(x)=x^3-2x as even, odd, or neither.
- f(-x) = (-x)^3 - 2(-x)-x^3 + 2x
- -f(x) = -(x^3 - 2x)-x^3 + 2x
- f(-x) = -f(x) for all xodd
Pro tip. For a quick test, compare f(-1) and f(1): equal nonzero values rule out odd; exact opposites are consistent with odd (and rule out even); any other pairing is neither. One pair never proves the identity for all x — it only screens.
If f is odd and f(3)=6, then f(-3) equals
- 6
- -6
- Cannot be determined
Odd means f(-x)=-f(x), so f(-3)=-6. Injectivity is not assumed — the sign rule alone fixes the answer.
5Modulus transformations
Modulus edits a graph in two standard ways. The graph of y=|f(x)| keeps positive values and reflects negative values above the axis — every dip below zero becomes a bump above. The graph of y=f(|x|) keeps the right half of f and mirrors it onto the left, producing an even curve even when f itself was not even.
Figure. Between -2 and 2 the dashed curve sits above the solid one — that is the negative part of f reflected upward. Outside that interval the two graphs coincide because f is already non-negative.
Modifying a parabola
For f(x)=x^2-4, compare f(0), |f(0)|, and f(|0|).
- f(0) = 0^2 - 4-4
- |f(0)| = |-4|4
- f(|0|) = f(0)-4
Pro tip. To count solutions of f(x)=k, draw y=f(x) and the horizontal line y=k and read intersections. Modulus on the outside changes how many times that line is met.
Starting from y=x^2-4, the graph of y=|x^2-4| differs from the original by
- Mirroring the right half onto the left
- Reflecting the part below the x-axis above it
- Shifting the vertex two units right
Outer modulus flips negative values upward. Mirroring the right half is what y=f(|x|) does, not y=|f(x)|.
6Graph shifts
The graph of y=f(x-h)+k moves the graph of y=f(x) right by h units and up by k. Signs trip people: a replacement like x-2 inside the input shifts the graph right, not left, because the input must increase to produce the same output. Combine shifts with stretches and reflections only after the base graph is clear.
Figure. The dashed parabola is the solid one moved two units right and one up. The vertex travels from (0,0) to (2,1) — every point on the curve makes the same journey.
Reading a shift
If f(x)=x^2, find the value of f(x-2)+1 at x=3.
- f(x-2)+1 at x=3 means f(1)+1substitute x=3
- f(1) = 1^21
- f(1)+12
Pro tip. To sketch y=f(x-2)+1, move every landmark of f two units right and one up — vertex, intercepts, and asymptotes all travel together.
7Quadratic maxima and minima
A quadratic y=ax^2+bx+c turns once, at x^*=-\frac{b}{2a}. If a>0 the parabola opens up and the vertex is a minimum; if the leading coefficient is negative, the parabola opens down and the vertex is a maximum. Substitute x^* directly into the quadratic — completing the square is optional, not required under time pressure.
Figure. The curve opens downward and peaks at x=2. Every other x-value gives a smaller y — that peak is the global maximum.
Maximum of a downward parabola
Find the maximum value of -2x^2+8x+5.
- a = -2,\ b = 8;\ a<0 so vertex is a maximumopens down
- x^* = -\frac{b}{2a} = -\frac{8}{-4}2
- y^* = -2(4)+8(2)+513
Pro tip. For a<0 the vertex value is the global maximum — no second check is needed. For positive x, the AM-GM bound x+\frac{k}{x}\ge2\sqrt{k} handles many single-variable minima without calculus.
The quadratic 3x^2-12x+1 has
- A maximum at x=2
- A minimum at x=2
- A minimum at x=-2
a=3>0 so the parabola opens up and the vertex is a minimum at x=-\frac{-12}{6}=2, not a maximum.
8Functional equations
CAT functional-equation items rarely ask for a full proof — they ask you to recognise the standard form and use one given value. If f(x+y)=f(x)+f(y), try f(x)=cx; if f(x+y)=f(x)f(y), try f(x)=a^x; if f(xy)=f(x)+f(y), try f(x)=\log x. One anchor value usually fixes the constant.
Figure. Match the given identity to a template, then one anchor value fixes the constant. Do not invent a new closed form mid-question.
Linear functional equation
A function satisfies f(x+y)=f(x)+f(y) for all reals and f(1)=5. Find f(4).
- Additive form \Rightarrow f(x)=cxlinear candidate
- f(1)=5 \Rightarrow c=5f(x)=5x
- f(4)=5\times420
Pro tip. Build up by repeated addition: f(4)=f(3)+f(1)=f(2)+f(1)+f(1), and each step uses the same rule. Exponential and logarithmic forms need positive inputs — check the domain before guessing.
Notes
- Function definition: A relation is a function when each input maps to exactly one output; an injective (one-to-one) function never repeats an output value.
- Composition: (f\circ g)(x)=f(g(x)), and composition is generally non-commutative, so f\circ g usually differs from g\circ f.
- Even and odd functions: f(-x)=f(x) marks an even function symmetric about the y-axis, while f(-x)=-f(x) marks an odd function symmetric about the origin.
- Modulus transformations: y=|f(x)| reflects any negative part of f above the x-axis, whereas y=f(|x|) mirrors the right half of the graph onto the left.
- Parabola extremum: y=ax^2+bx+c attains its vertex at x=-\frac{b}{2a}, giving a maximum when a<0 and a minimum when a>0.
Formulas
- Composition: (f\circ g)(x)=f(g(x))
- Inverse condition: f(f^{-1}(x))=x for a bijection
- Parabola vertex: x^*=-\frac{b}{2a},\ y^*=c-\frac{b^2}{4a}
- AM-GM optimum: x+\frac{k}{x}\ge2\sqrt{k} for x>0
- Graph shift: y=f(x-h)+k shifts the curve right by h and up by k
Exam traps & shortcuts
- To find a quadratic's maximum or minimum, substitute x=-\frac{b}{2a} straight into the function rather than completing the square.
- For the range of \frac{ax+b}{cx+d}, solve for x in terms of y and exclude the y that makes the resulting denominator zero.
- Count solutions of f(x)=k by sketching y=f(x) and the horizontal line y=k and reading off the intersections.
- For f(x+y)=f(x)f(y) guess f(x)=a^x; for f(xy)=f(x)+f(y) guess f(x)=\log x; for f(x+y)=f(x)+f(y) guess f(x)=cx.
Reference tables
Every line should be reconstructible from the concepts above, not merely recalled.
| Topic | Rule | Watch for |
|---|---|---|
| Composition | (f\circ g)(x)=f(g(x)) | Inner function runs first |
| Inverse | f(f^{-1}(x))=x | Needs bijection |
| Even / odd | f(-x)=f(x) / f(-x)=-f(x) | Most functions are neither |
| Modulus outside | y=|f(x)| | Reflects negative part up |
| Modulus inside | y=f(|x|) | Mirrors right half left |
| Shift | y=f(x-h)+k | x-h moves right, +k moves up |
| Vertex | x^*=-\frac{b}{2a},\ y^*=c-\frac{b^2}{4a} | a<0 gives a maximum |
Recap
Read only this the night before.
- Function
- One output per input. Injective means no repeated outputs.
- Composition
- Inside-out: (f\circ g)(x)=f(g(x)). Order usually matters.
- Inverse
- Swap x and y, solve, verify. Graph reflects across y=x.
- Symmetry
- Even: mirror on y-axis. Odd: rotate about origin.
- Modulus
- |f(x)| flips negatives up; f(|x|) mirrors the right half.
- Shifts
- f(x-h)+k: right h, up k. Sign of h is the trap.
- Quadratic
- Vertex at x=-b/2a. a<0 means maximum.
- Functional eq.
- Additive \to linear, multiplicative \to exponential, log form for f(xy).
Practise Functions, Graphs and Maxima-Minima
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 21 exam-style questions on this topic, with explanations
- A 10-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device