Last updated on July 16th, 2025
The non-negative distance of a number from zero is always provided by the absolute value function, represented as |x|. The function returns x if x is zero or positive, and βx 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) = β£xβ£; therefore, the output is -x if the input value π₯ is negative or zero, and f(x) is always non-negative if the input value π₯ is positive or zero.
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. X climbs when it is positive; it falls when it is negative.
Key characteristics of the absolute value function are graphically depicted and explained briefly.
Incapacity
Applying the absolute-value function twice produces the same outcome as applying it once alone.
|| x || = | x |
As β£π₯β£ is always non-negative, a second application has no effect.
For all real π₯, absolute values by definition quantify distance from zero, like
|x| ≥ 0
This characteristic is verified by the left plot's V-shaped curve's every point being at least above the x-axis.
A product's absolute value is equal to the product of its absolute values:
|xy|=|x| |y|.
This ensures consistent scaling of magnitudes regardless of sign.
If, and only if x = 0, then |x| = 0.
The function reaches zero exclusively at the vertex (0,0); it does not reach zero at any other point on the curve.
An even function satisfies symmetry (evenness) f(-x) = f(x). Now, here if
|-x| = |x|,
As shown by the equal values at x = -2 and x = 2, the graph is mirror-symmetric with regard to the y-axis.
For any real numbers π and π,
|a+b| ≤ |a|+|b|.
The triangle inequality is illustrated by the right-hand diagram, which displays points π = 1 and π = 2 on the number line. The direct distance from 0 to π + π = 3 (bottom arrow) is not greater than the sum of the distances from 0 to π and from π to π + π (top arrows).
Graphically, the absolute value function is shown as a "V"-shaped graph. The function has a basic π(π₯)=β£π₯β£; its graph includes a sharp vertex at the origin (0, 0). The graph follows the line π¦ = π₯ for values of π₯ > 0; it reflects across the y-axis, following the line π¦ = −π₯, for values of π₯ < 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 β£x − 3β£ = 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 β£Aβ£ = B—where B ≥ 0—can be represented as two distinct linear equations: A = B and A = −B. For instance, |x − 3β£ = 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 π§ = π + π, the absolute value—or modulus—of a complex number determines the distance from the origin to the point (a, b). It is provided by
|z| = √a2+b2,
So that the result is non-negative even if π or π is negative. In terms of geometry, β£π§β£ specifies how far π§ deviates from 0 + 0π, and in terms of algebra, it guarantees that complex number multiplication and division preserve magnitude
|zw| = |z| |w| and |z-1| = 1/|z|.
In calculus, the absolute value function is sometimes considered as a piecewise definition.
|x|={x, x ≥ 0}, {-x, x < 0},
For any real π₯, making it continuous but non-differentiable at π₯ = 0. As a function of distance from the origin, the sign function is d|x| / dx = 1 for x > 0 and -1 for x < 0. This behavior is necessary when evaluating limits, solving optimization issues, and integrating expressions involving β£π₯β£ since one must split integrals or take one-sided derivatives into account at the edge.
The sign of π₯ provides the derivative of the absolute value function f(x)= |xβ£ everywhere except at the origin, where it is absent. If,
d/dx|x| = {1, x > 0}, {-1, x < 0}.
The derivative is undefined at π₯ = 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, ′(π₯), is shown in the plot above. Open circles at π₯ = 0 indicate that the derivative is undefined at the edge, and it equals −1 for all π₯ < 0 and +1 for π₯ > 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 |= x for x ≥ 0 and |x| = -x for x < 0, its integral also needs to be assessed in sections. For instance, to incorporate ∫ |x| dx, the function has to be divided into
∫ |x|dx=-∞∫0 (-x) dx + 0∫∞ 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.
Useful illustrations of how absolute value measures distance, mistakes, and deviations in science, engineering, and finance.
Thorough guidelines to assist students in avoiding typical errors in graphing, solving, and interpreting absolute value expressions.
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 = 2/2 ⇒ x = 1
Then set 2x + 1 = -3, which will be 2x = -3 - 1
2x = -4 ⇒ x = - 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 = 0 will be the result.
Now, a different set x - 4 = -4will be produced, x = -4 + 4 = 8
Thus, the answer is x = 8, or x = 0
Solve |x|< 4.
-4 < x < 4
Convert into -4 < x < 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.