AMC 10 Step by Step

Functions

Function evaluation, composition, inverses, functional equations, graphs and transformations

25
primary-topic problems (1.9% of all)
9
more as a secondary topic
Where it appears
18
P1-10
1
P11-15
2
P16-20
4
P21-25

What you need to know

  • Composition. f(g(x))f(g(x)) means: apply gg first, then apply ff to the result. Work from the inside out. Order matters: f(g(x))f(g(x)) and g(f(x))g(f(x)) are usually different. Example: if f(x)=x+1f(x)=x+1 and g(x)=x2g(x)=x^2, then f(g(3))=f(9)=10f(g(3))=f(9)=10 but g(f(3))=g(4)=16g(f(3))=g(4)=16.
  • Inverse. f−1(b)=af^{-1}(b)=a means exactly the same as f(a)=bf(a)=b, so a question about f−1f^{-1} is really "solve f(a)=bf(a)=b". The graph of f−1f^{-1} is the graph of ff reflected across the line y=xy=x. Example: if f(x)=2x+3f(x)=2x+3, then f−1(11)f^{-1}(11) is the aa with 2a+3=112a+3=11, so f−1(11)=4f^{-1}(11)=4.
  • Moving graphs. Let cc be a positive number. f(x)+cf(x)+c moves the graph up by cc, and f(x−c)f(x-c) moves it right by cc. f(−x)f(-x) flips it across the yy-axis, and −f(x)-f(x) flips it across the xx-axis. f(cx)f(cx) squeezes it toward the yy-axis by a factor of cc (when c>1c>1). Example: y=(x−2)2+1y=(x-2)^2+1 is y=x2y=x^2 moved right 22 and up 11, so its lowest point is (2,1)(2,1).
  • Functional equations. These give a rule for ff instead of a formula. Solve them by substitution: plug in 00, 11, −1-1, and the "partner" input that appears in the equation (such as 1x\frac1x, 1−x1-x, or −x-x). This gives a small system of linear equations in the unknown values. Example: if f(x)−2f(−x)=xf(x)-2f(-x)=x for all xx, then x=0x=0 gives f(0)−2f(0)=0f(0)-2f(0)=0, so f(0)=0f(0)=0.
  • Symmetry. If f(x)=f(2a−x)f(x)=f(2a-x) for every xx, the graph is symmetric about the vertical line x=ax=a, because xx and 2a−x2a-x are the same distance from aa on opposite sides. Example: f(x)=(x−3)2f(x)=(x-3)^2 has f(x)=f(6−x)f(x)=f(6-x) (here a=3a=3), and indeed f(1)=f(5)=4f(1)=f(5)=4.
  • Piecewise and floor rules. If ff has different rules for different inputs, or uses the floor ⌊x⌋\lfloor x\rfloor (the greatest integer less than or equal to xx), split into cases by the input. Example: ⌊2.7⌋=2\lfloor2.7\rfloor=2 but ⌊−2.7⌋=−3\lfloor-2.7\rfloor=-3.

How AMC 10 tests it

  • Problems 3–10: evaluate a composition or an inverse from a formula or a small table, often working one step backwards.
  • Problems 8–15: "f(x)+2f ⁣(1x)=3xf(x)+2f\!\left(\frac1x\right)=3x; find f(2)f(2)". Substitute x=2x=2 and x=12x=\frac12 and solve the two equations in two unknowns.
  • Problems 10–18: rules such as f(x+y)=f(x)f(y)f(x+y)=f(x)f(y), where a clever choice of inputs pins down the value.
  • Graph questions: pick out the graph of y=f(x−1)+2y=f(x-1)+2, or count the solutions of f(f(x))=cf(f(x))=c from a sketch.
  • Problems 15–22: a function applied over and over, written f(n)(x)f^{(n)}(x) for nn applications (find when it starts repeating), or counting fixed points (inputs with f(x)=xf(x)=x).

Standard approaches

  1. Do compositions from the inside out. For inverses, solve f(a)=bf(a)=b instead of finding a formula for f−1f^{-1}.
  2. For a functional equation, substitute a specific value, then its partner value, and solve the system you get.
  3. For a function applied many times, work out f(f(x))f(f(x)) and f(f(f(x)))f(f(f(x))) with algebra and watch for xx to come back. Example: for f(x)=11−xf(x)=\frac1{1-x}, f(f(x))=x−1xf(f(x))=\frac{x-1}x and f(f(f(x)))=xf(f(f(x)))=x, so the values repeat every 33 steps.
  4. To move a graph, follow two or three landmark points, such as the intercepts or the vertex.
  5. To count solutions from a graph, count crossings with horizontal lines, one layer at a time. Example: for f(f(x))=cf(f(x))=c, first find every yy with f(y)=cf(y)=c, then count the solutions of f(x)=yf(x)=y for each one.

Worked example

The function ff satisfies f(x)+2f(1−x)=3x2f(x)+2f(1-x)=3x^2 for all real xx. What is f(3)f(3)?

(A) −3-3 (B) −1-1 (C) 11 (D) 33 (E) 55

Substitute x=3x=3. Since 1−3=−21-3=-2, this gives f(3)+2f(−2)=27f(3)+2f(-2)=27.

The equation pairs xx with 1−x1-x, so the partner of 33 is −2-2. Substitute x=−2x=-2. Since 1−(−2)=31-(-2)=3 and 3(−2)2=123(-2)^2=12, this gives f(−2)+2f(3)=12f(-2)+2f(3)=12, so f(−2)=12−2f(3)f(-2)=12-2f(3).

Put that into the first equation:
f(3)+2(12−2f(3))=27⟹24−3f(3)=27⟹f(3)=−1. f(3)+2\bigl(12-2f(3)\bigr)=27\quad\Longrightarrow\quad 24-3f(3)=27\quad\Longrightarrow\quad f(3)=-1.
Check: f(x)=2(1−x)2−x2f(x)=2(1-x)^2-x^2 satisfies the equation, and f(3)=2⋅4−9=−1f(3)=2\cdot4-9=-1.

The answer is (B) −1\boxed{\textbf{(B)}\ -1}.

Pitfalls

  • Composing in the wrong order (f(g(x))f(g(x)) versus g(f(x))g(f(x))).
  • Confusing f−1(x)f^{-1}(x) with 1f(x)\frac1{f(x)}. Example: for f(x)=2xf(x)=2x, f−1(x)=x2f^{-1}(x)=\frac x2, but 1f(x)=12x\frac1{f(x)}=\frac1{2x}.
  • Shifting the wrong way: f(x−2)f(x-2) moves the graph right, not left. Example: y=(x−2)2y=(x-2)^2 has its vertex at x=2x=2.
  • Using only one substituted equation. You need both the original and the partner equation to get rid of the extra unknown value.

Traps that recur

  • Bounding |f(800) − f(400)| by ½·400 = 200 directly (ignoring f(300) = f(900)) and answering 100. (2022 AMC 10B #24)
  • Applying the rule as if diamond were multiplication or subtraction, or computing 6 diamond x = 6x instead of 6/x. (2016 AMC 10A #23)
  • Summing 1 through 100 instead of 1 through 99 (exponent 5050), or reading f(2n) = n f(n) as f(2^n) = 2^n f(2^(n-1)). (2004 AMC 10A #24)
  • Plugging 3z in for x on the right side (as if f(x) = x^2 + x + 1), instead of first finding f(t) for a general input t. (2000 AMC 10 #24)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.

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