What Are Absolute Value Functions?
Absolute Value Functions return the distance of a number from zero, always as a value that is zero or positive. The parent function is $f(x) = |x|$, and it is defined in two pieces:
$$|x| = \begin{cases} x, & x \ge 0 \ -x, & x < 0 \end{cases}$$
The two branches meet the same rule: strip the sign and keep the size. For a positive input the output is the input itself; for a negative input the output is its opposite, which is positive. That is why $-x$ in the lower branch is not "a negative number", it is the positive partner of a negative $x$, since $-(-3) = 3$.
Because the definition switches at $x = 0$, an absolute value function is a piecewise function built from two straight lines. Written as one expression it is also $|x| = \sqrt{x^2}$, since the square erases the sign and the principal square root returns the non-negative root.
The three facts to carry forward:
Domain: all real numbers, because every number has a distance from zero.
Range: $y \ge 0$, because a distance is never negative.
Symmetry: $|-x| = |x|$, so $f$ is an even function and its graph is symmetric about the $y$-axis.
What Does The Graph Of An Absolute Value Function Look Like?
The graph of $f(x) = |x|$ is a V. The right branch is the line $y = x$ for $x \ge 0$; the left branch is the line $y = -x$ for $x < 0$. The two lines meet at the vertex $(0, 0)$, the lowest point of the graph and the exact spot where the function changes direction.
To the right of the vertex the graph climbs with slope $+1$.
To the left of the vertex it falls (reading left to right) with slope $-1$.
The vertex is a sharp corner, not a smooth turn. That corner is the whole story of the calculus later on.
The V never dips below the $x$-axis, which is the picture of the range $y \ge 0$. For the full graph-first treatment with worked plotting, see the algebra page on the absolute value function.
How Do You Transform An Absolute Value Function?
Every absolute value function you meet is the parent $|x|$ shifted, stretched, or flipped. The general form is:
$$f(x) = a,|x - h| + k$$
Each constant moves the V in a predictable way:
$h$ shifts the vertex horizontally to $x = h$ (right when $h > 0$, left when $h < 0$).
$k$ shifts the vertex vertically to $y = k$, so the vertex sits at $(h, k)$.
$a$ stretches the V (steeper arms when $|a| > 1$, wider when $0 < |a| < 1$) and, when $a < 0$, reflects it so the V opens downward.
So $f(x) = 2|x - 3| + 1$ has its vertex at $(3, 1)$, arms of slope $\pm 2$, and opens upward. For the general theory of shifting and scaling any graph, see transformations of functions.
Table: How each constant in $a,|x - h| + k$ changes the parent V.
Constant | Effect on the graph | Vertex moves to |
|---|---|---|
$h$ | Horizontal shift (right if $h>0$) | $x = h$ |
$k$ | Vertical shift (up if $k>0$) | $y = k$ |
$a$ (size) | Steeper if $ | a |
$a<0$ | Reflection: V opens downward | unchanged |
How Do You Solve An Absolute Value Equation?
Because $|x|$ hides the sign, an equation such as $|x - 3| = 5$ has two cases: the inside can be $+5$ or $-5$.
$$x - 3 = 5 \quad \text{or} \quad x - 3 = -5$$
Solving each gives $x = 8$ or $x = -2$. Both are genuine solutions, because the point $8$ and the point $-2$ each sit a distance $5$ from $3$ on the number line.
Inequalities split the same way, but the direction matters:
$|x - 3| < 5$ means the distance is less than $5$, giving the single band $-2 < x < 8$.
$|x - 3| > 5$ means the distance is greater than $5$, giving the two outer pieces $x < -2$ or $x > 8$.
A quick check keeps you honest: an equation like $|x| = -4$ has no solution, since a distance can never equal a negative number.
Is An Absolute Value Function Continuous And Differentiable?
Here the calculus begins, and it holds one clean surprise: $|x|$ is continuous everywhere but not differentiable at $x = 0$.
Continuity. The two branches meet at the vertex with no gap and no jump: as $x \to 0$ from either side, $|x| \to 0$, which equals $f(0)$. So
$$\lim_{x \to 0} |x| = 0 = f(0),$$
and the function is continuous for every real $x$. See continuity of a function for the full test.
Differentiability. Differentiability asks whether the graph has one well-defined tangent slope at a point. Everywhere except the vertex it does. But at $x = 0$ the two arms arrive with different slopes, so the one-sided derivatives disagree:
$$\lim_{h \to 0^-} \frac{|0 + h| - |0|}{h} = -1, \qquad \lim_{h \to 0^+} \frac{|0 + h| - |0|}{h} = +1$$
The left-hand slope is $-1$, the right-hand slope is $+1$, and because they are not equal the definition of the derivative limit does not exist at $0$. A sharp corner has no single tangent line. This is the textbook example that continuity does not force differentiability, the whole point of differentiability vs continuity: a function can be joined up yet still have a corner. The reverse always holds, since differentiability of a function at a point does guarantee continuity there.
What Is The Derivative Of An Absolute Value Function?
Away from the corner the derivative is simple, because each branch is a straight line.
For $x > 0$, $;|x| = x$, so $\dfrac{d}{dx}|x| = 1$.
For $x < 0$, $;|x| = -x$, so $\dfrac{d}{dx}|x| = -1$.
Both cases are captured by one formula, the signum (sign) function:
$$\frac{d}{dx}|x| = \frac{x}{|x|} = \operatorname{sgn}(x), \qquad x \neq 0$$
Read $\dfrac{x}{|x|}$ as "$+1$ when $x$ is positive, $-1$ when $x$ is negative", which matches the two slopes of the V. The derivative is deliberately left undefined at $x = 0$, exactly because the corner has no single slope. Geometrically, the derivative is the tangent slope, and the V hands you slope $+1$ or $-1$ and nothing in between.
Example 1: Differentiate $g(x) = |x - 3|$.
Shift does not change the slope pattern; the corner just moves to $x = 3$:
$$g'(x) = \frac{x - 3}{|x - 3|} = \operatorname{sgn}(x - 3), \qquad x \neq 3$$
Final answer: $g'(x) = +1$ for $x > 3$ and $g'(x) = -1$ for $x < 3$; undefined at $x = 3$.
What Is The Integral Of An Absolute Value Function?
Integrating $|x|$ recovers area, and the antiderivative can be written in one compact expression:
$$\int |x|,dx = \frac{x,|x|}{2} + C$$
This is worth checking by differentiating back, branch by branch:
For $x > 0$: $\dfrac{x|x|}{2} = \dfrac{x^2}{2}$, and $\dfrac{d}{dx}\left(\dfrac{x^2}{2}\right) = x = |x|$.
For $x < 0$: $\dfrac{x|x|}{2} = \dfrac{-x^2}{2}$, and $\dfrac{d}{dx}\left(\dfrac{-x^2}{2}\right) = -x = |x|$.
Both branches differentiate back to $|x|$, and the two pieces agree at $x = 0$, so the antiderivative is correct.
Example 2: Evaluate $\displaystyle\int_{-1}^{1} |x|,dx$.
The graph over $[-1, 1]$ is two right triangles, each with base $1$ and height $1$, so the area is $2 \times \tfrac{1}{2}(1)(1) = 1$. The antiderivative agrees:
$$\int_{-1}^{1} |x|,dx = \left[\frac{x|x|}{2}\right]_{-1}^{1} = \frac{(1)(1)}{2} - \frac{(-1)(1)}{2} = \frac{1}{2} + \frac{1}{2} = 1$$
Final answer: $\displaystyle\int_{-1}^{1} |x|,dx = 1$.
The geometric reading and the algebraic one match: the integral is the area under the V, and the two triangles add to exactly $1$.
Why Do Absolute Value Functions Behave This Way?
The V, the corner, and the range all trace back to one idea: an absolute value function reports distance, and distance discards direction.
The range is $y \ge 0$ because distance cannot be negative. No matter which way you step from zero, the count of how far is a non-negative number, so the graph never crosses below the axis.
The graph folds because sign is thrown away. Feeding in $-x$ gives the same output as $x$, so the left half of $y = x$ is reflected up into the V. The fold happens exactly at $x = 0$, where the sign flips.
The corner exists because two straight lines meet at an angle. Each branch is perfectly smooth on its own, but they arrive at the vertex from opposite tilts. A smooth curve would need the slopes to agree there; distance does not care, so the slopes clash and a corner forms.
That corner is why $|x|$ is the standard first example of "continuous but not differentiable". The function is glued together with no break, yet the glued point is a spike, not a smooth valley.
Who Shaped The Absolute Value Function And The Corner?
The vertical-bar notation and the careful idea of a "corner" both arrived in the 1800s, as mathematics moved from loose pictures to precise definitions.
Two more threads sit behind the notation:
Jean-Robert Argand (1768–1822, France) used the idea of the modulus, the distance of a complex number from the origin, which is the same "size without direction" that $|x|$ captures on the real line.
Karl Weierstrass (1815–1897, Germany) is also credited with popularising the vertical-bar notation $|x|$ for absolute value in his lectures, fixing the symbol that every textbook now uses.
Where Are Absolute Value Functions Used In The Real World?
Any time only the size of a gap matters, not its direction, an absolute value function is the natural tool.
Measurement error and tolerance: a part passes quality control when $|\text{measured} - \text{target}| \le \text{tolerance}$, since being oversize or undersize by the same amount is equally bad.
Distance and navigation: the gap between two positions on a line, $|a - b|$, is a distance regardless of which point is larger.
Physics: speed is the absolute value of velocity, $|v|$, when direction is set aside and only the rate matters.
Finance and data: the mean absolute deviation averages $|x_i - \bar{x}|$ to describe spread without letting positive and negative gaps cancel.
Signal processing: rectifying a signal takes $|x|$ to fold the negative half up, turning an alternating signal into a one-sided one.
One idea, "how far, not which way", runs quietly through engineering tolerances, physics, statistics, and electronics.
What Are The Most Common Mistakes With Absolute Value Functions?
These three errors account for most lost marks, and each matches a question real students ask on r/learnmath, CK-12, Brainly, and course error sheets.
Assuming $|x|$ is differentiable at $x = 0$.
Where it slips in:
A student sees that $|x|$ is continuous and smooth-looking away from the origin, then writes a derivative value at $x = 0$ as if the corner had a slope.
Don't do this:
Do not assign a derivative at the vertex. Continuity does not grant differentiability.
The correct way:
Check the one-sided slopes. The left gives $-1$, the right gives $+1$; since they disagree, $\dfrac{d}{dx}|x|$ is undefined at $0$, and defined as $\operatorname{sgn}(x)$ only for $x \neq 0$.
Writing $\sqrt{x^2} = x$ instead of $|x|$.
Where it slips in:
Simplifying a square root of a square, a student cancels the square and the root to get $x$, dropping the sign guard.
Don't do this:
Do not write $\sqrt{x^2} = x$. If $x$ is negative, the left side is positive while $x$ is not, so the two cannot be equal.
The correct way:
Keep the absolute value: $\sqrt{x^2} = |x|$. The principal square root is never negative, so the answer must be the non-negative version of $x$. The same guard gives $\sqrt{(x+2)^2} = |x + 2|$.
Misreading the $-x$ branch as "a negative answer".
Where it slips in:
Reading the definition $|x| = -x$ for $x < 0$, a student concludes the output is negative and panics that a distance came out below zero.
Don't do this:
Do not treat $-x$ as "negative". The minus sign is applied to an already-negative input.
The correct way:
Substitute a real number. For $x = -3$, the lower branch gives $-x = -(-3) = 3$, a positive output, exactly the distance of $-3$ from zero.
Practice Problems On Absolute Value Functions
Work each one, then check against the answer. Answers are verified.
State the vertex and range of $f(x) = |x + 4| - 2$.
(Answer: vertex $(-4, -2)$; range $y \ge -2$.)Solve $|2x - 1| = 7$.
(Answer: $2x - 1 = 7$ or $-7$, so $x = 4$ or $x = -3$.)Solve the inequality $|x + 5| \le 3$.
(Answer: $-3 \le x + 5 \le 3$, so $-8 \le x \le -2$.)Find $\dfrac{d}{dx}|x|$ at $x = -4$.
(Answer: $x < 0$, so the derivative is $-1$.)Evaluate $\displaystyle\int_{0}^{2} |x|,dx$.
(Answer: on $[0,2]$, $|x| = x$, so $\left[\tfrac{x^2}{2}\right]_0^2 = 2$.)Evaluate $\displaystyle\int_{-2}^{2} |x|,dx$.
(Answer: two triangles of area $2$ each, or $\left[\tfrac{x|x|}{2}\right]_{-2}^{2} = 2 + 2 = 4$.)
Where Can You Get Help With Absolute Value Functions?
If the corner at the vertex, the signum derivative, or the split into two cases is still slippery, a live tutor who can watch your working and point to the exact step is the fastest fix. Bhanzu's trainers teach absolute value functions from the distance picture up, so the graph, the derivative, and the integral feel like one idea rather than three rules to memorise.
High-school math tutoring for the algebra-to-calculus bridge where absolute value functions first turn tricky.
Online math tutors from India for one-to-one help fitted around your schedule.
GCSE maths tutoring for UK students meeting modulus graphs and equations.
Where Should You Go Next After Absolute Value Functions?
The corner in the V is a doorway into the heart of calculus, and several natural next steps open from here.
Differentiability of a function. Formalise what "having a derivative" means, using $|x|$ at $0$ as the first counterexample.
Definition of the derivative. See exactly how the one-sided limits split at the corner and why that kills the derivative.
Piecewise functions. Every absolute value function is one; this is the general machinery for functions defined in cases.
If your child is meeting absolute value functions for the first time, a live Bhanzu trainer teaches them from the distance picture up, so the graph and the calculus feel like one idea, in the Bhanzu high-school math program.
Was this article helpful?
Your feedback helps us write better content