Composition and Inverses

• Evaluate f(g(x)) by substituting g(x) into every x of f(x), and read a composition from a table of values from the inside out (h(f(g(2))))
• Find the inverse of a function by swapping x and y, then solving for y
• Evaluate a piecewise function by first selecting the correct piece for the given input
• Recognize that (f ∘ g)(x) is not the same as (g ∘ f)(x) in general
• Identify domain restrictions that determine which piece of a piecewise function applies

Function composition f(g(x)) means 'do g first, then apply f to the result.' To compute it, substitute the entire expression for g(x) wherever you see x in f(x). For example, if f(x) = 2x + 3 and g(x) = x², then f(g(x)) = f(x²) = 2(x²) + 3 = 2x² + 3. Order matters: g(f(x)) = g(2x + 3) = (2x + 3)², which is different. The ACT loves to test which order the problem is asking for. Read 'f of g of x' as 'g is inside, f is outside.' When the functions are given as a table, composition is lookup from the inside out: for h(f(g(2))), find g(2) in the table, use that output as the input to f, then use f's output as the input to h. To solve h(f(g(a))) = 1 from a table, run it backward from the outside: find the x where h(x) = 1, then the x where f(x) equals that number, then the x where g(x) equals that one; that last x is a. An inverse function f⁻¹(x) undoes f(x): if f(3) = 7, then f⁻¹(7) = 3. To find an inverse algebraically: (1) write y = f(x), (2) swap x and y, (3) solve the new equation for y, (4) write the result as f⁻¹(x). Example: f(x) = 3x − 5. Swap: x = 3y − 5. Solve: y = (x + 5)/3. So f⁻¹(x) = (x + 5)/3. Check: f(f⁻¹(x)) should give back x. The same swap works for roots and fractions: for f(x) = ∛(x + 2) − 5, swap to x = ∛(y + 2) − 5, add 5, cube both sides, and f⁻¹(x) = (x + 5)³ − 2; the inverse undoes f's steps in reverse order. Not every function has an inverse. Only one-to-one functions do. Piecewise functions are defined with different rules over different parts of their domain. Example: f(x) = { x² if x < 0; 2x + 1 if x ≥ 0 }. To evaluate f(−3), use the first piece (because −3 < 0): f(−3) = 9. To evaluate f(4), use the second piece: f(4) = 9. Always check which interval the input falls into before computing. The ACT trap on piecewise questions is applying the wrong piece. Students rush and use the rule that's written first without checking the condition. When two piecewise functions are composed, check the conditions again at every step: in f(g(−2)), pick g's piece for −2 (a boundary value belongs to the piece whose condition has the equals sign), then pick f's piece for whatever g(−2) turns out to be. On graphs, a piecewise function often has a jump or corner at the boundary point. The graph of f⁻¹ is the reflection of the graph of f over the line y = x, so if (a, b) is on f, then (b, a) is on f⁻¹. Hard composition questions can run backward: given f(x) and f(g(x)), treat g(x) as the unknown input to f and solve for it. For a combined transformation y = a·f(x − h) + k, a point (p, q) on f moves to (p + h, a·q + k): multiply the y-value by a first, then add k.

• For f(g(x)), compute g(x) first, then plug that entire expression into f. Write the intermediate step. Do not skip it.
• To find an inverse, swap x and y, then solve for y. This is the only method worth memorizing.
• For piecewise functions, check the inequality first: which piece does the input satisfy? Then apply that rule.
• f(g(x)) ≠ g(f(x)) almost always. If a question gives both, compute both separately.
• Verify an inverse by checking f(f⁻¹(x)) = x. If it does not simplify to x, you made an algebra error.