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308 LearnersLast updated on October 21, 2025

The absolute value function represents the non-negative distance of a number from zero. The function returns x if x is zero or positive, and –x if x is negative. Its "V"-shaped graph, which has a vertex at the origin (0, 0), is frequently used to show deviation and distance.
Without taking direction into account, the absolute value function calculates a number's separation from zero on the number line. It helps express magnitude regardless of sign because it always yields a non-negative result.
It is stated as \(f(x) = ∣x∣\);
Therefore, the output is x if the input value x is positive or zero, and x if the input value x is negative. In both cases, f(x) is always non-negative.
For example, \(|0| = 0, |-3| = 3\), and \(|5| = 5\). With its vertex at the origin (0, 0), the graph of the absolute value function has an unusual V form.
Here, the function grows as x becomes positive and declines as x becomes negative.
Key characteristics of the absolute value function are graphically depicted and explained briefly.
Applying the absolute-value function twice produces the same outcome as applying it once alone.
\(|| x || = | x |\)
As ∣𝑥∣ is always non-negative, a second application has no effect.
Graphically, the absolute value function is shown as a "V"-shaped graph. The function has a basic \(𝑓(𝑥)=∣𝑥∣\); its graph includes a sharp vertex at the origin (0, 0).
The graph follows the line \(𝑦 = 𝑥\) for values of \(𝑥 > 0\); it reflects across the y-axis, following the line \(𝑦 = −𝑥\), for values of \(𝑥 < 0\).
The function exhibits even symmetry about the y-axis. Whether x is positive or negative, the graph shows that the absolute value always returns non-negative outcomes based on both sides of the vertex.
An absolute value equation is an equation including a variable within absolute value bars, such as \(∣x − 3∣ = 5\).
The key to solving it is realizing that the expression inside the bars can be either positive or negative, and that it will still provide the same non-negative outcome.
Any absolute value equation of the type \(∣A∣ = B\), where B ≥ 0, can be represented as two distinct linear equations:
A = B and A = −B.
For instance, \(|x − 3∣ = 5\) can be broken apart into \(x − 3 = 5\) and \(x − 3 = −5\), therefore producing \(x = 8\) and \(x = −2\).
This methodical process guarantees the discovery of all legitimate equation solutions.
With the formula \(z = a + bi\), the absolute value of a complex number determines the distance from the origin to the point (a, b). It is provided as,
\(|z| = \sqrt{a^2 + b^2} \),
So that the result is non-negative even if a or b is negative.
In terms of geometry, ∣𝑧∣ specifies how far 𝑧 deviates from 0 + 0𝑖, and in terms of algebra, it guarantees that complex number multiplication and division preserve magnitude
\(|zw| = |z| \, |w| \\ |z - 1| = \frac{1}{|z|} \).
In calculus, the absolute value function is often expressed as a piecewise function.
|x|={x, x ≥ 0}, {-x, x < 0},
For any real 𝑥, making it continuous but non-differentiable at \(𝑥 = 0\). As a function of distance from the origin, the sign function is,
\(\frac{d}{dx} |x| = \begin{cases} 1 & \text{if } x > 0, \\ -1 & \text{if } x < 0. \end{cases}\)
This behavior is necessary when evaluating limits, solving optimization issues, and integrating expressions involving ∣𝑥∣ since one must split integrals or take one-sided derivatives into account at the edge.
When f(x) is given as f(x) = |x-1|, the graph comes out as,
The sign of 𝑥 provides the derivative of the absolute value function f(x)= |x∣ everywhere except at the origin, where it is absent. If,
\(\frac {d}{dx}|x|\) = {1, x > 0}, {-1, x < 0}.
The derivative is undefined at 𝑥 = 0 because the graph has a sharp "corner" that prevents the definition of a single tangent line there.
The fact that the absolute-value function is continuous but not differentiable at its vertex is reflected in this behavior.
The derivative of the absolute value function, ′(𝑥), is shown in the plot above. Open circles at \(𝑥 = 0\) indicate that the derivative is undefined at the edge, and it equals −1 for all \(𝑥 < 0\) and +1 for \(𝑥 > 0\).
When integrating an absolute value function, the area under its graph is calculated, which frequently necessitates segmenting the function according to the point at which the equation inside the absolute value changes sign.
Given that the absolute value function is defined piecewise, usually as,
\(|x| = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0 \end{cases}\)
Its integral also needs to be assessed in sections. For instance, to incorporate \(∫ |x| dx\), the function has to be divided into
\(\int |x| \, dx = \int_{-\infty}^{0} (-x) \, dx + \int_{0}^{\infty} x \, dx \)
Because of the nature of absolute value, this method guarantees that the total area is calculated accurately and is always non-negative.
Including such features is very helpful in applications involving total distance or scale, independent of direction.
Here are some student-friendly tips and tricks to master the absolute value function:
Plot a few examples such as:
∣3∣=3 (3 units to the right of 0)
∣−3∣=3 (3 units to the left of 0)
Remember that the absolute value can represent distance between two numbers. For example, ∣x−y∣ represents the distance between x and y on the number line.
Learn the shape of the absolute value graph. The graph of y = |x| is a V-shape. The vertex is at (0, 0) and its slope is 1 for x > 0 and −1 for x < 0.
Check for extraneous solutions. When squaring or manipulating equations with absolute values, always check both solutions in the original equation — sometimes one might not work.
Thorough guidelines to assist students in avoiding typical errors in graphing, solving, and interpreting absolute value expressions.
Useful illustrations of how absolute value measures distance, mistakes, and deviations in science, engineering, and finance.
Temperature differences: Absolute value is used to find temperature changes without worrying about “up” or “down.”
Example:
If it’s 5°C in the morning and −3°C at night:
Temperature difference \(= |5 - (-3)| = |8| = 8°C\)
Solve |x|=5.
\(x = 5, -5\)
Determine that the \(|x| = 5\) has split into,
\(x = 5\ or\ x = -5\)
Therefore, the answer is \(x = 5,-5\)
Solve |x -3| = 7
\(x = 10, -4\)
First, we will set \(x - 3 = 7\),
Which will be \(x = 7 + 3 = 10\)
Then \(x - 3 = -7\),
Which will be \(x = -7 + 3 = -4\)
We obtain \(x = 10\) and \(x = -4\),
Thus, the answer is \(x = 10,-4\)
Solve |2x + 1| = 3
\(x = 1, -2\)
Set \(2x + 1 = 3\), which will be \(2x = 3 - 1 = 2\)
\(x = \frac {2}2\)
\(x = 1\)
Then set \(2x + 1 = -3\), which will be \(2x = -3 - 1\)
\(2x = -4\)
\(x = \frac {-4}2\)
\(x = -2\)
Therefore, the answer is x = 1 -2.
Solve |x-4| + 2 = 6
\(x = 8, 0\)
Remove 2 from the expression \(|x - 4| = 6 - 2 = 4\)
Therefore, it \(|x - 4| = 4\) will be the absolute.
If \(x - 4 = 4\),
Then \(x = 4 + 4 = 8\) will be the result.
Now, a different set x - 4 = -4
\(x = -4 + 4 = 0\)
Thus, the answer is \(x = 8\), or \(x = 0\)
Solve |x|< 4.
\(-4 < x < 4\)
This converts to \(-4 < x < 4\),
Which means x lies between -4 and 4.
This puts x between -4, and 4
Therefore, it \(-4 < x < 4\) is the answer.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






