Pre-Calculus

Functions and Their Graphs: Transformations, Inverse, and Composite Functions

Pre-Calculus

✓By the end of this lesson students will be able to identify and apply transformations (translations, reflections, stretches, and compressions) to the graphs of parent functions.
✓By the end of this lesson students will be able to determine if a function is one-to-one and find the equation of its inverse function.
✓By the end of this lesson students will be able to evaluate and determine the domain of composite functions.
✓By the end of this lesson students will be able to understand the relationship between the domain and range of a function and its inverse.

Key concepts

Transformations of Functions

Transformations alter the position, size, or orientation of a function's graph. These include translations (shifts), reflections (flips), and stretches/compressions (dilations). Given a parent function y = f(x), common transformations are:\n- Vertical Translation: y = f(x) + c (up c units), y = f(x) - c (down c units)\n- Horizontal Translation: y = f(x - c) (right c units), y = f(x + c) (left c units)\n- Reflection: y = -f(x) (across x-axis), y = f(-x) (across y-axis)\n- Vertical Stretch/Compression: y = c * f(x) (stretch if c > 1, compression if 0 < c < 1)\n- Horizontal Stretch/Compression: y = f(c * x) (compression if c > 1, stretch if 0 < c < 1)\nWhen multiple transformations are applied, the order generally follows: reflections and stretches/compressions first, then translations.

Inverse Functions

An inverse function, denoted f⁻¹(x), 'undoes' the action of the original function f(x). For f⁻¹(x) to exist, the function f(x) must be one-to-one, meaning each output corresponds to exactly one input. This can be tested graphically using the Horizontal Line Test. If any horizontal line intersects the graph more than once, the function is not one-to-one.\nTo find the inverse function algebraically:\n1. Replace f(x) with y.\n2. Swap x and y in the equation.\n3. Solve the new equation for y.\n4. Replace y with f⁻¹(x).\nThe domain of f(x) is the range of f⁻¹(x), and the range of f(x) is the domain of f⁻¹(x). The graph of f⁻¹(x) is a reflection of the graph of f(x) across the line y = x.

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x

Composite Functions

A composite function is formed when one function is substituted into another function. The notation (f o g)(x) means f(g(x)), where the output of the inner function g(x) becomes the input for the outer function f(x).\nTo evaluate (f o g)(x):\n1. Evaluate the inner function g(x) first.\n2. Use the result as the input for the outer function f(x).\nThe domain of (f o g)(x) consists of all x-values in the domain of g(x) such that g(x) is in the domain of f(x).

(f o g)(x) = f(g(x))

Key facts to remember

1Horizontal transformations (inside the function) often behave counter-intuitively (e.g., x+c shifts left).
2A function must pass the Horizontal Line Test (be one-to-one) to have an inverse function.
3The graph of an inverse function f⁻¹(x) is a reflection of the graph of f(x) across the line y = x.
4The domain of f(x) is the range of f⁻¹(x), and the range of f(x) is the domain of f⁻¹(x).
5The domain of a composite function f(g(x)) includes all x in the domain of g such that g(x) is in the domain of f.
6Order of transformations matters: generally, reflections and stretches/compressions are applied before translations.

Worked examples

Example 1

Describe the sequence of transformations applied to the graph of the parent function f(x) = |x| to obtain the graph of g(x) = -3|x - 2| + 4.

IIdentify the parent function: f(x) = |x|.

IIIdentify the transformations in the order they should be applied (reflections/stretches/compressions first, then translations):

III1. The factor of '3' outside the absolute value indicates a vertical stretch by a factor of 3. (y = 3|x|)

IV2. The negative sign outside the absolute value indicates a reflection across the x-axis. (y = -3|x|)

V3. The '(x - 2)' inside the absolute value indicates a horizontal translation 2 units to the right. (y = -3|x - 2|)

VI4. The '+ 4' outside the absolute value indicates a vertical translation 4 units up. (y = -3|x - 2| + 4)

Answer

The graph of g(x) is obtained from f(x) = |x| by:\n1. Vertically stretching by a factor of 3.\n2. Reflecting across the x-axis.\n3. Translating 2 units to the right.\n4. Translating 4 units up.

The order of transformations is crucial. Generally, reflections and stretches/compressions are applied before translations.

Example 2

Find the inverse function f⁻¹(x) for f(x) = (2x + 1) / (x - 3). State the domain and range of both f(x) and f⁻¹(x).

I1. Replace f(x) with y: y = (2x + 1) / (x - 3).

II2. Swap x and y: x = (2y + 1) / (y - 3).

III3. Solve for y:

IV x(y - 3) = 2y + 1

V xy - 3x = 2y + 1

VI xy - 2y = 3x + 1

VII y(x - 2) = 3x + 1

VIII y = (3x + 1) / (x - 2)

94. Replace y with f⁻¹(x): f⁻¹(x) = (3x + 1) / (x - 2).

105. Determine the domain and range of f(x):

11 For f(x), the denominator cannot be zero, so x - 3 ≠ 0, which means x ≠ 3. Domain of f(x): (-∞, 3) U (3, ∞).

12 To find the range of f(x), we can find the domain of its inverse. Alternatively, for rational functions of the form (ax+b)/(cx+d), the horizontal asymptote is y = a/c. So, y ≠ 2/1 = 2. Range of f(x): (-∞, 2) U (2, ∞).

136. Determine the domain and range of f⁻¹(x):

14 For f⁻¹(x), the denominator cannot be zero, so x - 2 ≠ 0, which means x ≠ 2. Domain of f⁻¹(x): (-∞, 2) U (2, ∞).

15 The range of f⁻¹(x) is the domain of f(x). Range of f⁻¹(x): (-∞, 3) U (3, ∞).

Answer

f⁻¹(x) = (3x + 1) / (x - 2)\nDomain of f(x): (-∞, 3) U (3, ∞)\nRange of f(x): (-∞, 2) U (2, ∞)\nDomain of f⁻¹(x): (-∞, 2) U (2, ∞)\nRange of f⁻¹(x): (-∞, 3) U (3, ∞)

Notice that the domain of f(x) is the range of f⁻¹(x), and the range of f(x) is the domain of f⁻¹(x).

Example 3

Given f(x) = sqrt(x + 5) and g(x) = x² - 1. Find (f o g)(x) and state its domain.

I1. Find the expression for (f o g)(x) = f(g(x)):

II Substitute g(x) into f(x): f(x² - 1) = sqrt((x² - 1) + 5)

III Simplify: f(g(x)) = sqrt(x² + 4).

IV2. Determine the domain of (f o g)(x):

V The domain of g(x) = x² - 1 is all real numbers, (-∞, ∞).

VI For f(g(x)) = sqrt(x² + 4) to be defined, the expression under the square root must be non-negative:

VII x² + 4 ≥ 0.

VIII Since x² is always greater than or equal to 0, x² + 4 will always be greater than or equal to 4. Therefore, x² + 4 is always non-negative for all real numbers x.

9 Thus, the domain of (f o g)(x) is all real numbers.

Answer

(f o g)(x) = sqrt(x² + 4)\nDomain of (f o g)(x): (-∞, ∞)

When finding the domain of a composite function, you must consider both the domain of the inner function and the domain restrictions imposed by the outer function on the output of the inner function.

Common mistakes

✗Applying transformations in the incorrect order, especially confusing the order of stretches/compressions with translations.
✗Incorrectly interpreting horizontal transformations (e.g., thinking f(x+c) shifts right instead of left).
✗Assuming every function has an inverse without checking if it's one-to-one, or failing to restrict the domain to make it one-to-one.
✗Forgetting to consider the domain of the inner function when determining the domain of a composite function.
✗Algebraic errors when solving for y after swapping x and y to find an inverse function.

Exam tips

★Always verify your inverse function by checking if f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
★When describing transformations, be precise with direction (left/right, up/down) and type (stretch/compression, reflection).
★Pay close attention to domain restrictions, especially for functions involving square roots (radicand ≥ 0) and rational functions (denominator ≠ 0).
★Practice with a variety of parent functions (e.g., absolute value, quadratic, square root, cubic) to master transformations.

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