Modulus Functions

The modulus (or absolute-value) function strips the sign from a real number, returning its non-negative magnitude. Although the idea is elementary, the modulus is one of the most powerful pieces of algebraic notation you will meet at A-Level: it lets you sketch reflections of graphs, solve equations with multiple cases, write inequalities as compact distance statements, and bound errors in numerical work. This lesson covers $|x|$∣x∣ and the piecewise definition, the two graph transformations $y = |f(x)|$y=∣f(x)∣ and $y = f(|x|)$y=f(∣x∣), the standard methods for solving modulus equations and inequalities, the geometric "distance from a" interpretation, and the role of $|x - a|$∣x−a∣ in error analysis.

---

Spec mapping

Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.8 covers the modulus function, including the notation $|x|$∣x∣, and use relations such as $|a| = |b| \Leftrightarrow a^2 = b^2$∣a∣=∣b∣⇔a2=b2 and $|x - a| < b \Leftrightarrow a - b < x < a + b$∣x−a∣<b⇔a−b<x<a+b in the course of solving equations and inequalities (refer to the official specification document for exact wording). Modulus content is examined on Paper 1 (Pure Mathematics) and is synoptic with Section 2.7 (graphs and proportion, especially graph transformations), Section 2.4 (inequalities, where modulus inequalities sit) and Section 7 (differentiation, where the non-differentiability of $|x|$∣x∣ at $x = 0$x=0 is a stock example of a continuous-but-not-differentiable function). A typical Paper 1 modulus question carries 6–10 marks and combines sketching, equation-solving and a "find values of $k$k" twist.

---

Definition of $|x|$∣x∣ and the piecewise form

The modulus (or absolute value) of a real number $x$x, written $|x|$∣x∣, is its distance from zero on the number line. Distance is non-negative, so $|x| \geq 0$∣x∣≥0 for every real $x$x, with equality only at $x = 0$x=0.

Formally, the modulus is defined piecewise:

$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$∣x∣={x−x​if x≥0if x<0​

So $|7| = 7$∣7∣=7, $|-3| = -(-3) = 3$∣−3∣=−(−3)=3, and $|0| = 0$∣0∣=0. Note that on the lower branch the formula is $-x$−x (which is positive when $x$x is negative) — not the symbol $x$x with a minus sign tacked on as decoration.

Key algebraic identities

Identity	Meaning
$	x
$	x
$	-x
$	xy
$	x
$\sqrt{x^2} =	x
$	a + b

The identity $|x|^2 = x^2$∣x∣2=x2 is the engine behind almost every modulus-equation technique on the spec: it lets you replace a stubborn $|f(x)|$∣f(x)∣ with $\sqrt{f(x)^2}$f(x)2​ and then square both sides.

The triangle inequality $|a + b| \leq |a| + |b|$∣a+b∣≤∣a∣+∣b∣ is foundational. It captures the idea that the direct distance between two points is never longer than the path that goes via a detour. A-Level work uses it sparingly, but it underpins the entire theory of metric spaces at university (see Going further below).

---

Sketching $y = |f(x)|$y=∣f(x)∣ — reflect the negative part above the x-axis

Given the graph of $y = f(x)$y=f(x), the graph of $y = |f(x)|$y=∣f(x)∣ is obtained by:

1. Keep the part of $f(x)$f(x) that is on or above the x-axis unchanged.
2. Reflect the part of $f(x)$f(x) that is below the x-axis in the x-axis (so it now lies above).

Why? Because $|f(x)| = f(x)$∣f(x)∣=f(x) when $f(x) \geq 0$f(x)≥0, and $|f(x)| = -f(x)$∣f(x)∣=−f(x) when $f(x) < 0$f(x)<0. The transformation $y \to -y$y→−y is reflection in the x-axis, applied only on the intervals where $f(x)$f(x) was negative.

The points where $f(x) = 0$f(x)=0 are the fold lines — the x-intercepts become "corner" points (kinks) in $y = |f(x)|$y=∣f(x)∣. The graph is continuous through them but not smooth; its gradient changes sign abruptly.

Example: Sketch $y = |x^2 - 4|$y=∣x2−4∣.

Start with $y = x^2 - 4$y=x2−4: a parabola, vertex at $(0, -4)$(0,−4), x-intercepts at $x = \pm 2$x=±2. Between $x = -2$x=−2 and $x = 2$x=2 the parabola dips below the x-axis. Reflect that dip above:

• Outside $[-2, 2]$[−2,2]: graph is $y = x^2 - 4$y=x2−4 (unchanged, opens upward, both branches above the axis).
• Inside $[-2, 2]$[−2,2]: graph is $y = -(x^2 - 4) = 4 - x^2$y=−(x2−4)=4−x2 (an upside-down parabola from $(-2, 0)$(−2,0) through $(0, 4)$(0,4) to $(2, 0)$(2,0)).

The result is a "W"-shape with corners at $(\pm 2, 0)$(±2,0) and a local maximum at $(0, 4)$(0,4).

---

Sketching $y = f(|x|)$y=f(∣x∣) — replace negative-x with a reflection of positive-x

Given the graph of $y = f(x)$y=f(x), the graph of $y = f(|x|)$y=f(∣x∣) is obtained by:

1. Keep the part of $f(x)$f(x) for $x \geq 0$x≥0 unchanged.
2. Discard the part of $f(x)$f(x) for $x < 0$x<0.
3. Reflect the kept (right-hand) portion in the y-axis to fill in the left half.

Why? Because $|x| = x$∣x∣=x when $x \geq 0$x≥0 and $|x| = -x$∣x∣=−x when $x < 0$x<0. So for negative $x$x, $f(|x|) = f(-x)$f(∣x∣)=f(−x), which is the reflection of $f$f in the y-axis evaluated at $x$x.

A defining feature: the graph of $y = f(|x|)$y=f(∣x∣) is always symmetric about the y-axis — it is automatically an even function, regardless of what $f$f was originally.

Example: Sketch $y = (|x| - 1)(|x| - 3)$y=(∣x∣−1)(∣x∣−3).

Start with $y = (x - 1)(x - 3)$y=(x−1)(x−3): a parabola with x-intercepts at $x = 1, 3$x=1,3, vertex at $(2, -1)$(2,−1), y-intercept $y = 3$y=3.

Take the right half ($x \geq 0$x≥0): the parabola from $(0, 3)$(0,3) down through $(1, 0)$(1,0) to vertex $(2, -1)$(2,−1) then up through $(3, 0)$(3,0).

Reflect in the y-axis to fill the left half: a mirror copy from $(-3, 0)$(−3,0) up to $(-2, -1)$(−2,−1) down to $(-1, 0)$(−1,0) then up to $(0, 3)$(0,3).

The result has x-intercepts at $\pm 1, \pm 3$±1,±3, two minima at $(\pm 2, -1)$(±2,−1), and a y-intercept of $3$3. It is symmetric about the y-axis.

$y = |f(x)|$y=∣f(x)∣ vs $y = f(|x|)$y=f(∣x∣) comparison

| Feature | $y = |f(x)|$y=∣f(x)∣ | $y = f(|x|)$y=f(∣x∣) | |---------|-----------------|-----------------| | What gets reflected | The part below the x-axis | The part for negative $x$x is discarded; positive-x portion is reflected in the y-axis | | Axis of reflection | x-axis | y-axis | | Effect on values | Output is non-negative everywhere | Input is treated as if non-negative | | Symmetry | Inherits any symmetry $f$f already had | Always symmetric about the y-axis | | Differentiability | Loses smoothness at every x-intercept of $f$f | Loses smoothness at $x = 0$x=0 (unless $f'(0) = 0$f′(0)=0) | | Range | $[0, \infty)$[0,∞) subset of original range | Same as range of $f$f on $[0, \infty)$[0,∞) |

A useful reading rule: the modulus on the outside flips below to above; the modulus on the inside mirrors right onto left.

---

Solving modulus equations $|f(x)| = a$∣f(x)∣=a

There are two standard methods. Both are on the spec, both score full marks, and the "right" choice depends on context.

Method 1 — case analysis

$|f(x)| = a$∣f(x)∣=a (with $a \geq 0$a≥0) splits into two equations:

$f(x) = a \quad \text{or} \quad f(x) = -a$f(x)=aorf(x)=−a

Solve each, then combine.

Worked example 1: Solve $|2x - 5| = 7$∣2x−5∣=7.

Case 1: $2x - 5 = 7 \Rightarrow 2x = 12 \Rightarrow x = 6$2x−5=7⇒2x=12⇒x=6.

Case 2: $2x - 5 = -7 \Rightarrow 2x = -2 \Rightarrow x = -1$2x−5=−7⇒2x=−2⇒x=−1.

Both solutions check ($|2(6) - 5| = |7| = 7$∣2(6)−5∣=∣7∣=7, $|2(-1) - 5| = |-7| = 7$∣2(−1)−5∣=∣−7∣=7). So $x = 6$x=6 or $x = -1$x=−1.

Method 2 — squaring both sides

Since $|f(x)|^2 = f(x)^2$∣f(x)∣2=f(x)2, the equation $|f(x)| = a$∣f(x)∣=a (for $a \geq 0$a≥0) is equivalent to $f(x)^2 = a^2$f(x)2=a2, which can be rearranged as $f(x)^2 - a^2 = 0$f(x)2−a2=0 and factorised as a difference of squares:

$(f(x) - a)(f(x) + a) = 0$(f(x)−a)(f(x)+a)=0

This gives the same two cases as Method 1, but as a single algebraic step.

Modulus-equation case-analysis table

Equation	Cases produced	Validity check needed?
$	f(x)	= a$,$,a \geq 0$
$	f(x)	= a$,$,a < 0$
$	f(x)	= g(x)$
$	f(x)	=

The second row is a pure-spec trap: if a question asks you to solve $|2x + 1| = -3$∣2x+1∣=−3, the answer is "no solutions" because the modulus is never negative. Spotting this and writing it explicitly earns a B1 mark.

---

Solving $|f(x)| = |g(x)|$∣f(x)∣=∣g(x)∣ — squaring is rigorous here

When both sides are inside a modulus, squaring is not just convenient but rigorously equivalent: there are no extraneous roots to discard. The reason is that for any reals $u, v$u,v:

$|u| = |v| \quad \Leftrightarrow \quad u^2 = v^2$∣u∣=∣v∣⇔u2=v2

Both sides of this equivalence are non-negative (modulus is non-negative; squares are non-negative), so squaring preserves equality in both directions. Contrast this with the asymmetric case $|f(x)| = g(x)$∣f(x)∣=g(x), where $g(x)$g(x) can be negative — squaring there can introduce solutions that don't satisfy the original equation.

Worked example 2: Solve $|2x - 1| = |x + 4|$∣2x−1∣=∣x+4∣.

Square both sides:

$(2x - 1)^2 = (x + 4)^2$(2x−1)2=(x+4)2

Expand:

$4x^2 - 4x + 1 = x^2 + 8x + 16$4x2−4x+1=x2+8x+16

Rearrange:

$3x^2 - 12x - 15 = 0$3x2−12x−15=0

Divide by 3:

$x^2 - 4x - 5 = 0$x2−4x−5=0

Factorise: $(x - 5)(x + 1) = 0$(x−5)(x+1)=0, so $x = 5$x=5 or $x = -1$x=−1.

Both are genuine solutions: $|2(5) - 1| = 9 = |5 + 4|$∣2(5)−1∣=9=∣5+4∣ and $|2(-1) - 1| = 3 = |-1 + 4|$∣2(−1)−1∣=3=∣−1+4∣. No checking step was needed because squaring is rigorous when both sides are moduli — but it is good practice to write the verification anyway.

Alternative method: case analysis. Write $2x - 1 = x + 4$2x−1=x+4 giving $x = 5$x=5, or $2x - 1 = -(x + 4)$2x−1=−(x+4) giving $3x = -3$3x=−3, $x = -1$x=−1. Same answers.

---

Solving $|f(x)| < g(x)$∣f(x)∣<g(x) inequalities

Modulus inequalities are where students lose the most marks, because the case analysis is more delicate than for equations and the squaring shortcut is not always valid.

The "distance form" $|x - a| < b$∣x−a∣<b

For a constant $a$a and $b > 0$b>0:

$|x - a| < b \quad \Leftrightarrow \quad a - b < x < a + b$∣x−a∣<b⇔a−b<x<a+b

This reads naturally as "$x$x is within distance $b$b of $a$a" — an open interval of length $2b$2b centred at $a$a. This compact form is one of the spec-named identities.

Similarly:

$|x - a| > b \quad \Leftrightarrow \quad x < a - b \text{ or } x > a + b$∣x−a∣>b⇔x<a−b or x>a+b

(an open exterior of the same interval).

Worked example 3: Solve $|3x - 2| < 5$∣3x−2∣<5.

Using the distance form: $|3x - 2| < 5 \Leftrightarrow -5 < 3x - 2 < 5 \Leftrightarrow -3 < 3x < 7 \Leftrightarrow -1 < x < \tfrac{7}{3}$∣3x−2∣<5⇔−5<3x−2<5⇔−3<3x<7⇔−1<x<37​.

So $x \in (-1, \tfrac{7}{3})$x∈(−1,37​).

General $|f(x)| < g(x)$∣f(x)∣<g(x) — case analysis

When $g(x)$g(x) is itself a function (not a constant), the cleanest method is case analysis based on the sign of $f(x)$f(x):

• Where $f(x) \geq 0$f(x)≥0: the inequality becomes $f(x) < g(x)$f(x)<g(x).
• Where $f(x) < 0$f(x)<0: the inequality becomes $-f(x) < g(x)$−f(x)<g(x), i.e. $f(x) > -g(x)$f(x)>−g(x).

Combining: the solution set is where $-g(x) < f(x) < g(x)$−g(x)<f(x)<g(x), and $g(x) > 0$g(x)>0 (because if $g(x) \leq 0$g(x)≤0 then no $f(x)$f(x) has $|f(x)| < g(x)$∣f(x)∣<g(x)).

Squaring inequalities — a warning

You can square the inequality $|f(x)| < g(x)$∣f(x)∣<g(x) to give $f(x)^2 < g(x)^2$f(x)2<g(x)2 only when $g(x) \geq 0$g(x)≥0. If $g(x)$g(x) might be negative, squaring reverses the truth value: the LHS is non-negative, the RHS is non-negative, but you have lost the information that $g(x) \geq 0$g(x)≥0 was needed for the inequality to hold at all. Many candidates lose marks here by squaring blindly without checking the sign of $g(x)$g(x).

---

The $|x - a|$∣x−a∣ as "distance from a" interpretation

The modulus has a beautiful geometric reading: $|x - a|$∣x−a∣ is the distance from $x$x to $a$a on the number line. From this, every modulus statement becomes a distance statement.

Algebraic form	Geometric meaning
$	x - a
$	x - a
$	x - a
$	x - a
$	x - a

The midpoint reading lets you skip algebra entirely: solving $|x - 3| = |x - 11|$∣x−3∣=∣x−11∣ by hand involves squaring and a quadratic, but geometrically the answer is just "the midpoint of 3 and 11", i.e. $x = 7$x=7. A* candidates who recognise the geometry write down the answer in one line.

This perspective is the launchpad for metric spaces at university, where $d(x, y) = |x - y|$d(x,y)=∣x−y∣ is the standard metric on $\mathbb{R}$R and the modulus's familiar properties (non-negativity, symmetry, triangle inequality) become the defining axioms of distance.

---

Modulus and absolute-value error bounds

In numerical work and applied mathematics, the modulus is the natural language of error. If the true value of a quantity is $x$x and your measured or computed estimate is $\hat{x}$x^, the absolute error is $|x - \hat{x}|$∣x−x^∣ — a non-negative number, regardless of whether you over- or under-estimated.

Statements like "the iteration converges to within $10^{-6}$10−6" are formalised as $|x_n - L| < 10^{-6}$∣xn​−L∣<10−6 for all sufficiently large $n$n. This is exactly the structure that Karl Weierstrass formalised in the epsilon–delta definition of limit: $\lim_{x \to a} f(x) = L$limx→a​f(x)=L means for every $\varepsilon > 0$ε>0 there exists $\delta > 0$δ>0 such that $|x - a| < \delta$∣x−a∣<δ implies $|f(x) - L| < \varepsilon$∣f(x)−L∣<ε. The whole edifice of rigorous calculus is built on modulus inequalities.

At A-Level you only need the introductory version: an answer is correct to "within $\pm 0.5$±0.5" means $|x - x_{\text{true}}| \leq 0.5$∣x−xtrue​∣≤0.5. But it is worth knowing where this language leads.

---

Worked example 4 — find $k$k such that $|f(x)| = k$∣f(x)∣=k has exactly two solutions

This question type tests whether you genuinely understand the V-shape geometry of a modulus graph rather than mechanically applying a procedure.

Question: The function $f$f is defined by $f(x) = |x^2 - 6x + 5|$f(x)=∣x2−6x+5∣ for $x \in \mathbb{R}$x∈R. Find the values of the constant $k$k for which the equation $f(x) = k$f(x)=k has exactly two solutions.

Solution. Step 1: sketch $y = x^2 - 6x + 5$y=x2−6x+5. Factorise: $(x - 1)(x - 5)$(x−1)(x−5), so x-intercepts at $x = 1, 5$x=1,5. Vertex at $x = 3$x=3, where $y = 9 - 18 + 5 = -4$y=9−18+5=−4. Below the x-axis on $1 < x < 5$1<x<5.