What is the domain of the modulus function ?

The domain of the modulus function is the set of all real values .

What is the range of the modulus function ?

The range of the modulus function is the set of all real non-negative values .

What can be said about the continuity of a modulus function at the origin ?

A modulus function is continuous at the origin. However, it is not differentiable .

True or False? When sketching a reflection of a modulus function the graphs should look smooth at the points where the original graph has been reflected .

False. When sketching a reflection of a modulus function the graphs should not look smooth at the points where the original graph has been reflected, they should look sharp .

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Modulus Functions & Further Transformations (DP IB Analysis & Approaches (AA))

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FrontModulus Functions
What is the modulus function?
BackModulus Functions
The modulus function, $f (x) = |x|$ , also known as the absolute value function, gives the magnitude of a number regardless of its sign .
It can be defined as
$|x| = \sqrt{x^{2}}$
or $|x| = \{\begin{cases} x & x \geq 0 \\ - x & x < 0 \end{cases}$

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What is the modulus function?
The modulus function, $f (x) = |x|$ , also known as the absolute value function, gives the magnitude of a number regardless of its sign .
It can be defined as
- $|x| = \sqrt{x^{2}}$
- or $|x| = \{\begin{cases} x & x \geq 0 \\ - x & x < 0 \end{cases}$
What is the domain of the modulus function?
The domain of the modulus function is the set of all real values.
What is the range of the modulus function?
The range of the modulus function is the set of all real non-negative values.
What is the y-intercept of the graph $y = | x |$ ?
What us the y-intercept of the graph $y = | x |$ is (0, 0).
True or False?
The graph $y = | x |$ has a single root.
True.
The graph $y = | x |$ has a single root at (0, 0).
What is the equation of the line of symmetry of the graph $y = | x |$ ?
The line of symmetry of the graph $y = | x |$ has equation $x = 0$ .
It is symmetrical about the y-axis.
How many roots of a graph of the form $y = a | x + p | + q$ have?
A graph of the form $y = a | x + p | + q$ can have 0, 1 or 2 roots.
True or False?
A graph of the form $y = a | x + p | + q$ has a vertex at $(- p, q)$ .
True.
A graph of the form $y = a | x + p | + q$ has a vertex at $(- p, q)$ .
True or False?
$|p - x| \neq |x - p|$
False.
$|p - x| = |x - p|$
For example, $|4 - x| = |x - 4|$ .
What can be said about the continuity of a modulus function at the origin?
A modulus function is continuous at the origin.
However, it is not differentiable.
How can the graph of the modulus function $y = |f (x)|$ be sketched?
The graph of the modulus function $y = |f (x)|$ can be sketched by:
- keeping any parts of the graph $y = f (x)$ that already lie above the x-axis, $y \geq 0$ ,
- and reflecting any parts that lie below the x-axis, across the x-axis.
How is the graph of the modulus function $y = f (|x|)$ sketched?
The graph of the modulus function $y = f (|x|)$ can be sketched by:
- keeping any parts of the graph $y = f (x)$ that already lie on the right-hand side of the y-axis, $x \geq 0$ ,
- and reflecting this section across the y-axis.
What is the difference between $y = |f (x)|$ and $y = f (|x|)$ ?
The difference between $y = |f (x)|$ and $y = f (|x|)$ is that:
- The graph of $y = |f (x)|$ never goes below the x-axis and it does not have to have any lines of symmetry.
- The graph of $y = f (|x|)$ is always symmetrical about the y-axis and it can go below the x-axis.
True or False?
When sketching a reflection of a modulus function the graphs should look smooth at the points where the original graph has been reflected.
False.
When sketching a reflection of a modulus function the graphs should not look smooth at the points where the original graph has been reflected, they should look sharp.
In what order should transformations that are applied outside a modulus function be carried out?
Transformations that are applied outside a modulus function should follow the same order as the order of operations.
- For a function of the form $y = |a f (x) + b|$ , deal with the $a$ then the $b$ then the modulus.
- For a function of the form $y = a |f (x)| + b$ , deal with the modulus then the $a$ then the $b$ .
In what order should transformations that are applied inside a modulus function be carried out?
Transformations that are applied inside a modulus function should follow the reverse order to the order of operations.
- For a function of the form $y = f (|a x + b|)$ , deal with modulus then the $b$ then the $a$ .
- For a function of the form $y = f (a |x| + b)$ , deal with the $b$ then the $a$ then the modulus.
How do you find the modulus of an equation, e.g. $f (x) = x - 8$ ?
You can find the modulus of an equation by keeping all positive outputs of the function and changing the sign of all negative outputs of the function, $|f (x)| = \{\begin{cases} f (x) & f (x) \geq 0 \\ - f (x) & f (x) < 0 \end{cases}$
E.g. $f (x) = x - 8$ , $f (8) = 0$ , so when $x \geq 8$ , $|f (x)| = f (x)$ , when $x < 8$ , $|f (x)| = - f (x)$ .
How can a modulus equation, $|f (x)| = g (x)$ , be solved graphically?
A modulus equation, $|f (x)| = g (x)$ , can be solved graphically by:
- drawing $y = |f (x)|$ and $y = g (x)$ into your GDC,
- and finding the x-coordinates of the points of intersection.
How can a modulus equation, $|f (x)| = g (x)$ , be solved analytically?
A modulus equation, $|f (x)| = g (x)$ , be solved analytically by:
- forming two equations ( $f (x) = g (x)$ and $f (x) = - g (x)$ ),
- solving both equations,
- and checking that the solutions work in the original equation.
What two equations need to be formed to solve the modulus equation $|\frac{2 x + 7}{3}| = x - 5$ .
The two equations need to be formed to solve the modulus equation $|\frac{2 x + 7}{3}| = x - 5$ , are:
- $\frac{2 x + 7}{3} = x - 5$
- $\frac{2 x + 7}{3} = - (x - 5)$
True or False?
The modulus inequality, $|f (x)| < g (x)$ can be solved by solving the two pairs of inequalities:
- $f (x) > g (x)$ , when $f (x) \geq 0$
- $f (x) < - g (x)$ , when $f (x) \leq 0$
False.
The modulus inequality, $|f (x)| < g (x)$ can not be solved by solving the two pairs of inequalities:
- $f (x) > g (x)$ , when $f (x) \geq 0$
- $f (x) < - g (x)$ , when $f (x) \leq 0$
These inequalities will provide the solution to $|f (x)| > g (x)$ .
To solve $|f (x)| < g (x)$ , solve the inequalities:
- $f (x) < g (x)$ , when $f (x) \geq 0$
- $f (x) > - g (x)$ , when $f (x) \leq 0$
What happens to the coordinates of a graph that undergoes the reciprocal transformation $\frac{1}{f (x)}$ ?
When a graph undergoes the reciprocal transformation $\frac{1}{f (x)}$ , the coordinates $(x, y)$ become $(x, \frac{1}{y})$ , where $y \neq 0$ .
I.e., the x-coordinates remain the same and the y-coordinates become their reciprocals.
True or False?
When a graph undergoes the reciprocal transformation $y = \frac{1}{f (x)}$ , any points on the graph that lie on the line y = 1 or the line y = -1 , stay the same.
True.
When a graph undergoes the reciprocal transformation $y = \frac{1}{f (x)}$ , any points on the graph that lie on the line y = 1 or the line y = -1 , stay the same.
What happens to a y-intercept $(0, c)$ of $y = f (x)$ in the reciprocal graph $y = \frac{1}{f (x)}$ ?
The y-intercept of the reciprocal graph $y = \frac{1}{f (x)}$ , is at $(0, \frac{1}{c})$ .
How does a root of $y = f (x)$ transform in the reciprocal graph $y = \frac{1}{f (x)}$ ?
A root of $y = f (x)$ becomes a vertical asymptote in the reciprocal graph $y = \frac{1}{f (x)}$ .
True or False?
If $y = f (x)$ is increasing, then $y = \frac{1}{f (x)}$ is also increasing.
False.
If $y = f (x)$ is increasing, then $y = \frac{1}{f (x)}$ is decreasing.
What happens to a horizontal asymptote $y = k$ $(k \neq 0)$ of $f (x)$ in the reciprocal graph?
If $f (x)$ has a horizontal asymptote $y = k$ $(k \neq 0)$ , then the reciprocal graph $y = \frac{1}{f (x)}$ has a horizontal asymptote at $y = \frac{1}{k}$ .
In the square transformation $y = {[f (x)]}^{2}$ , what happens to points below the x-axis?
In the square transformation $y = {[f (x)]}^{2}$ :
- points below the x-axis are reflected above the x-axis,
- and the distance between the x-axis and the point is squared.
If $(x_{1}, y_{1})$ is a point on $y = f (x)$ , what is the corresponding point on $y = {[f (x)]}^{2}$ ?
If $(x_{1}, y_{1})$ is a point on $y = f (x)$ , then the corresponding point on $y = {[f (x)]}^{2}$ is $(x_{1}, {y_{1}}^{2})$ .
How does a root of $y = f (x)$ transform in the square graph $y = {[f (x)]}^{2}$ ?
A root of $y = f (x)$ becomes a root and turning point in $y = {[f (x)]}^{2}$ .
True or False?
The graph of $y = {[f (x)]}^{2}$ can have negative y-values.
False.
The graph of $y = {[f (x)]}^{2}$ can never go below the x-axis.

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Modulus Functions & Further Transformations (DP IB Analysis & Approaches (AA))

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1. Number & Algebra

Number & Algebra Toolkit

Exponentials & Logs

Sequences & Series

Simple Proof & Reasoning

Proof by Induction & Contradiction

Binomial Theorem

Permutations & Combinations

Complex Numbers

Further Complex Numbers

Systems of Linear Equations

2. Functions

Linear Functions & Graphs

Quadratic Functions & Graphs

Functions Toolkit

Other Functions & Graphs

Reciprocal & Rational Functions

Transformations of Graphs

Polynomial Functions

Inequalities

Modulus Functions & Further Transformations

3. Geometry & Trigonometry

Geometry Toolkit

Geometry of 3D Shapes

Trigonometry

The Unit Circle & Exact Values

Trigonometric Functions & Graphs

Trigonometric Equations & Identities

Inverse & Reciprocal Trigonometric Functions

Trigonometric Proof & Equation Strategies

Vector Properties

Vector Equations of Lines

Vector Planes

4. Statistics & Probability

Statistics Toolkit

Correlation & Regression

Probability

Discrete Random Variables

Binomial Distribution

Normal Distribution

Continuous Random Variables

5. Calculus

Differentiation

Techniques & Applications of Differentiation

Integration

Techniques & Applications of Integration

Optimisation

Kinematics

Basic Limits & Continuity

Further Differentiation

Further Integration

Differential Equations

Maclaurin Series

Limits using l'Hôpital's Rule & Maclaurin Series