IB Maths AA functions: composites, inverses, transformations and logs

What functions covers at SL and HL

This follows the current Analysis and Approaches guide (first assessment 2021). A new maths course is first assessed in 2029; exams in 2027 and 2028 use this content.

The key ideas

A function maps every input in its domain to exactly one output. The domain is the set of allowed inputs, and the range is the set of outputs that actually occur. Most domain restrictions come from three places: you cannot divide by zero, you cannot take the square root of a negative number, and you cannot take the logarithm of zero or a negative number.

The composite (f o g)(x) means f(g(x)): apply g first, then f. Order matters, and (f o g)(x) is usually different from (g o f)(x). The inverse f^-1 undoes f. It only exists if f is one to one, its graph is the reflection of y = f(x) in the line y = x, and the domain of f^-1 is the range of f.

Transformations follow a rule worth memorising: changes inside the bracket act on x and work the opposite way to how they look (f(x - 2) moves the graph 2 to the right), while changes outside act on y as they look (f(x) + 1 moves it up 1). Exponentials and logarithms are inverses of each other, so e^(ln x) = x and ln(e^x) = x, and the laws of logarithms turn products into sums.

Worked example 1: composite and inverse functions

1. Let f(x) = 2x + 3 and g(x) = x^2 - 1. Then (f o g)(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 + 1.
2. And (g o f)(x) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 8. Check with x = 1: g(f(1)) = g(5) = 24, and 4 + 12 + 8 = 24.
3. Now find the inverse of h(x) = (2x + 1)/(x - 3), x not equal to 3. Write y = (2x + 1)/(x - 3), so y(x - 3) = 2x + 1.
4. Collect the x terms: xy - 2x = 3y + 1, so x(y - 2) = 3y + 1 and x = (3y + 1)/(y - 2).
5. Swap the letters: h^-1(x) = (3x + 1)/(x - 2), x not equal to 2. Check: h(4) = 9/1 = 9 and h^-1(9) = 28/7 = 4. The excluded value 2 is the horizontal asymptote of h, which is why the range of h is all reals except 2.

Worked example 2: a combined transformation

1. The point (1, 4) lies on y = f(x). Find its image on y = 3f(x - 2) + 1.
2. Inside the bracket, x - 2 is a translation 2 units to the right: (1, 4) becomes (3, 4).
3. Outside, the factor 3 is a vertical stretch, scale factor 3: (3, 4) becomes (3, 12).
4. Then + 1 is a translation 1 unit up: (3, 12) becomes (3, 13).
5. Check by substitution: on the new graph at x = 3, y = 3f(3 - 2) + 1 = 3f(1) + 1 = 3 × 4 + 1 = 13. The order matters: stretching after adding 1 would wrongly give 15.

Worked example 3: an exponential equation and the discriminant

1. Solve 3^(2x) - 4 × 3^x + 3 = 0. Notice that 3^(2x) = (3^x)^2, so let u = 3^x.
2. The equation becomes u^2 - 4u + 3 = 0, which factorises to (u - 1)(u - 3) = 0, so u = 1 or u = 3.
3. Then 3^x = 1 gives x = 0, and 3^x = 3 gives x = 1. Check x = 1: 9 - 12 + 3 = 0. Check x = 0: 1 - 4 + 3 = 0.
4. Discriminant: for which k does kx^2 + 4x + 1 = 0 have two distinct real roots? Need b^2 - 4ac > 0, so 16 - 4k > 0 and k < 4.
5. Also k cannot be 0, or the equation is not a quadratic. Answer: k < 4, k not equal to 0. Check with k = 3: 16 - 12 = 4 > 0, two roots.

Common mistakes that cost marks

• Doing a composite in the wrong order: (f o g)(x) means g first.
• Writing an inverse without its domain, or giving the domain of f instead of its range.
• Moving a graph left for f(x - 2). Changes inside the bracket go the opposite way.
• Applying a vertical stretch and translation in the wrong order.
• Splitting logs illegally: ln(a + b) is not ln a + ln b.
• Keeping solutions that make a logarithm undefined. Always check log equations against the domain.
• Forgetting that the leading coefficient cannot be zero in a discriminant question.

Exam technique and how a tutor helps

Functions questions often open a long problem: find the inverse, state its domain, then sketch both graphs. Sketches are marked on key features, so label intercepts, asymptotes and end points, and draw asymptotes as dashed lines. On Paper 2, use the GDC to find intersections and check a sketch, but still write down what you found and how.

When a question says 'hence', use the previous part. When it asks for an exact answer, keep logs and fractions, such as x = ln 5/2, rather than decimals.

In one-to-one lessons a tutor gets the student to explain each transformation out loud while sketching on the shared whiteboard, which exposes the inside-versus-outside confusion quickly. Lessons then mix short drills on inverses and log laws with past-paper questions where functions set up a calculus or probability problem. HL students get regular practice with polynomial roots and modulus inequalities, which Paper 1 likes to test.

Self-check: can you do these?

• If f(x) = x^2 and g(x) = x + 3, find f(g(-1)). (Answer: 4)
• Find the inverse of f(x) = e^x - 2 and state its domain. (Answer: f^-1(x) = ln(x + 2), x > -2)
• State the asymptotes of y = (x + 1)/(2x - 4). (Answer: x = 2 and y = 1/2)
• Solve log_2(x) + log_2(x - 2) = 3. (Answer: x = 4; x = -2 is rejected)
• Write y = 2x^2 - 8x + 5 in vertex form and give the vertex. (Answer: y = 2(x - 2)^2 - 3, vertex (2, -3))