Composite Functions fg(x), Inverse Functions f⁻¹(x) & Graph Transformations
On the GCSE Maths Higher Tier, a function f(x) is an algebraic rule that maps an input x to a single output; combining two functions in sequence creates a composite function fg(x), and reversing the mapping creates the inverse function f⁻¹(x). In particular, in the composite function fg(x), you always apply the inner function g(x) first (working from right to left) and substitute it into the outer function f! This topic also covers graph transformations, where changes outside the bracket affect the y-coordinates vertically as expected, while changes inside the bracket affect the x-coordinates horizontally in the opposite direction.
- Function Notation & Evaluating f(a)
- f(x) represents the output of function f for input x. To evaluate f(3), substitute x = 3 into the expression for f(x). If told that f(k) = 15, set the algebraic expression for f(k) equal to 15 and solve the equation for k.
- Composite Functions: fg(x) and ff(x)
- fg(x) means f(g(x)): apply function g first, then apply function f to the output of g. For algebraic expressions, replace every x in f(x) with the entire bracketed expression (g(x)). Similarly, ff(x) (or f²(x)) means f(f(x)): substitute the expression for f(x) back into f(x) itself.
- Inverse Functions: f⁻¹(x)
- The inverse function f⁻¹(x) reverses the effect of f(x) (so if f(2) = 9, then f⁻¹(9) = 2, and ff⁻¹(x) = x). To find f⁻¹(x), write y = f(x), rearrange the formula to make x the subject in terms of y, and then replace y with x to state f⁻¹(x). Note: f⁻¹(x) does not mean 1 / f(x)!
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Function Notation & Evaluating f(a)
- f(x) represents the output of function f for input x. To evaluate f(3), substitute x = 3 into the expression for f(x). If told that f(k) = 15, set the algebraic expression for f(k) equal to 15 and solve the equation for k.
- Composite Functions: fg(x) and ff(x)
- fg(x) means f(g(x)): apply function g first, then apply function f to the output of g. For algebraic expressions, replace every x in f(x) with the entire bracketed expression (g(x)). Similarly, ff(x) (or f²(x)) means f(f(x)): substitute the expression for f(x) back into f(x) itself.
- Inverse Functions: f⁻¹(x)
- The inverse function f⁻¹(x) reverses the effect of f(x) (so if f(2) = 9, then f⁻¹(9) = 2, and ff⁻¹(x) = x). To find f⁻¹(x), write y = f(x), rearrange the formula to make x the subject in terms of y, and then replace y with x to state f⁻¹(x). Note: f⁻¹(x) does not mean 1 / f(x)!
- Vertical & Horizontal Graph Translations
- Given the graph of y = f(x) with a turning point at (x₀, y₀): 1. y = f(x) + a is a vertical translation by vector (0, a), moving every point UP by a units to (x₀, y₀ + a). 2. y = f(x + a) is a horizontal translation by vector (-a, 0), moving every point left by a units to (x₀ - a, y₀). 3. y = f(x - a) is a horizontal translation by vector (a, 0), moving every point right by a units to (x₀ + a, y₀).
- Graph Reflections in the Coordinate Axes
- Given the graph of y = f(x) with a point at (x₀, y₀): 1. y = -f(x) negates all y-coordinates (outside the bracket), which is a reflection in the x-axis, mapping (x₀, y₀) to (x₀, -y₀). 2. y = f(-x) negates all x-coordinates (inside the bracket), which is a reflection in the y-axis, mapping (x₀, y₀) to (-x₀, y₀).
- Transforming Trigonometric Graphs (sin x, cos x, tan x)
- For y = sin(x) (which passes through (0°, 0), has a maximum at (90°, 1), crosses at (180°, 0), and has a minimum at (270°, -1)): y = sin(x - 30°) shifts the entire wave 30° to the right (so maximum is at (120°, 1)), and y = -cos(x) reflects the cosine wave in the x-axis (so it starts at (0°, -1) and has a maximum at (180°, 1)).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Given f(x) = 3x - 5 and g(x) = x², work out fg(-2) and gf(-2).
Find the inverse function f⁻¹(x) when f(x) = (4x - 7) / 5.
If a curve y = f(x) has a minimum turning point at (3, -5), state the coordinates of the turning point on: (a) y = f(x - 2) + 4, (b) y = -f(x), and (c) y = f(-x).
Describe fully the single geometrical transformation that maps the graph of y = f(x) onto the graph of y = f(x + 6).
Describe fully the single geometrical transformation that maps the graph of y = f(x) onto the graph of y = -f(x).
How can you solve the equation f⁻¹(x) = 8 quickly without even finding the algebraic formula for f⁻¹(x)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
The functions f and g are defined as f(x) = 2x - 3 and g(x) = x² + 4x. (a) Find an expression for fg(x) in its simplest form. (b) Solve the equation gf(x) = 12. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) To find fg(x), substitute g(x) = x² + 4x in place of x in f(x): fg(x) = 2(x² + 4x) - 3 = 2x² + 8x - 3 [1 mark]. 2. (b) To find gf(x), substitute f(x) = 2x - 3 in place of every x in g(x): gf(x) = (2x - 3)² + 4(2x - 3) [1 mark]. 3. Expand and simplify gf(x): (4x² - 12x + 9) + (8x - 12) = 4x² - 4x - 3 [1 mark]. 4. Set gf(x) = 12 and rearrange to equal zero: 4x² - 4x - 3 = 12 -> 4x² - 4x - 15 = 0 [1 mark]. 5. Factorise and solve: (2x - 5)(2x + 3) = 0 -> x = 2.5 (or 5/2) or x = -1.5 (or -3/2) [1 mark].
- Q2
The function h is defined as h(x) = (3x + 5) / (x - 4) for x ≠ 4. Find an expression for the inverse function h⁻¹(x), and state the value of x that must be excluded from the domain of h⁻¹(x). (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Write y = (3x + 5) / (x - 4) and multiply both sides by (x - 4): y(x - 4) = 3x + 5 -> xy - 4y = 3x + 5 [1 mark]. 2. Rearrange so all terms containing x are on the left-hand side and all other terms are on the right: xy - 3x = 4y + 5 [1 mark]. 3. Factorise x out on the left and divide by (y - 3): x(y - 3) = 4y + 5 -> x = (4y + 5) / (y - 3) [1 mark]. 4. Replace y with x to write h⁻¹(x) = (4x + 5) / (x - 3), and exclude x = 3 (since division by zero is undefined) [1 mark].
- Q3
The curve C has equation y = x² - 6x + 11. (a) By completing the square, write y in the form (x - p)² + q and state the coordinates of the minimum turning point of C. (b) The curve C is reflected in the y-axis and then translated by the vector (0, -5) to form curve D. Find the coordinates of the turning point of curve D. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Halve the coefficient of x (-6 ÷ 2 = -3) to complete the square: x² - 6x + 11 = (x - 3)² - (-3)² + 11 = (x - 3)² - 9 + 11 = (x - 3)² + 2 [1 mark]. 2. Therefore, the minimum turning point of curve C is at (3, 2) [1 mark]. 3. (b) Reflecting (3, 2) in the y-axis (y = f(-x)) negates the x-coordinate, moving the turning point to (-3, 2) [1 mark]. 4. Translating (-3, 2) by the vector (0, -5) subtracts 5 from the y-coordinate, giving the turning point of curve D as (-3, -3) [1 mark].
- Q4
The graph of y = cos(x) for 0° ≤ x ≤ 360° has a maximum point at (0°, 1), crosses the x-axis at (90°, 0) and (270°, 0), and has a minimum point at (180°, -1). Find the coordinates of the minimum point and the x-intercepts for the transformed graph y = -cos(x + 45°) in the interval 0° ≤ x ≤ 360°. (Higher Tier) [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Inside the bracket, (x + 45°) translates the graph 45° to the left (subtracts 45° from all x-coordinates) [1 mark]. 2. Outside the bracket, the minus sign in -cos(.) reflects the graph in the x-axis (multiplies all y-coordinates by -1, so old maximums become minimums and vice versa) [1 mark]. 3. The original maximum at (360°, 1) shifts left 45° to (315°, 1) and reflects in the x-axis to become the minimum point at (315°, -1) (while the x-intercepts at 90° and 270° shift left 45° to (45°, 0) and (225°, 0)) [1 mark].
Common Exam Questions (4)
In the composite function fg(x), which function do you apply first?
Always work from the inside out (from right to left): fg(x) means f(g(x)), so you apply g first and substitute the entire expression for g(x) in place of every 'x' inside the function f! For example, if f(x) = x² + 1 and g(x) = 3x - 2, then fg(x) = (3x - 2)² + 1, whereas gf(x) = 3(x² + 1) - 2. Note that fg(x) and gf(x) are almost never the same.
What are the 4 standard steps to find an inverse function f⁻¹(x)?
1. Write the function as y = . instead of f(x) = . 2. Swap the letters x and y everywhere in the equation (if x appears twice, such as in a fraction, you can rearrange to make the original x the subject first, then swap at the end). 3. Rearrange the equation to make the new y the subject. 4. Replace y with f⁻¹(x).
How do you find f⁻¹(x) when x appears in both the numerator and denominator of a fraction, like f(x) = (2x + 1) / (x - 3)?
When x appears twice in a fraction: 1. Write y = (2x + 1) / (x - 3) and multiply up by the denominator: y(x - 3) = 2x + 1. 2. Expand the bracket: xy - 3y = 2x + 1. 3. Collect all terms containing x on the left-hand side and all other terms on the right: xy - 2x = 3y + 1. 4. Factorise x out on the left: x(y - 2) = 3y + 1 -> divide by (y - 2): x = (3y + 1) / (y - 2). 5. Swap x and y to write f⁻¹(x) = (3x + 1) / (x - 2).
How do you remember the rules for graph transformations f(x) + a, f(x + a), -f(x), and f(-x)?
Use the 'Outside = Vertical (Normal), Inside = Horizontal (Opposite)' rule. 1. Outside the bracket changes y (vertical): y = f(x) + a translates the graph UP by a units [vector (0, a)]; y = -f(x) multiplies y by -1, so it reflects the graph in the X-axis. 2. Inside the bracket changes x (horizontal, opposite): y = f(x + a) translates the graph left by a units [vector (-a, 0)]; y = f(x - a) translates right by a units [vector (a, 0)]; y = f(-x) multiplies x by -1, so it reflects the graph in the Y-axis.
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