Understand · explore · practise
Sketch y = |f(x)| and y = f(|x|), transform points, and check domains and asymptotes. Compare both rules interactively and practise original graph questions.
Before you startThe modulus function, graph coordinates, domain and range
01 / Two different operations
In |f(x)|, evaluate f first, then make its output nonnegative. In f(|x|), first make the input nonnegative, then evaluate f. The second operation can still produce negative outputs.
Output modulus: x → f → f(x) → modulus
Input modulus: x → modulus → |x| → f
With f(x) = 2x − 2 and x = −1, |f(−1)| = |−4| = 4, but f(|−1|) = f(1) = 0. The positions of the bars matter.
02 / Sketch y = |f(x)|
Keep every point on or above the x-axis. Reflect every point below it upwards across the x-axis. The input stays the same: (a, b) becomes (a, |b|).
The x-intercepts stay at the same input values. A smooth crossing may become a corner after reflection. The domain stays the same, because taking the modulus of an already defined real output imposes no new restriction.
f(x) = x² − 4
Its zeros are ±2 and its vertex is (0, −4).
|f(x)| = 4 − x² for −2 ≤ x ≤ 2
Reflect the portion below the axis. The vertex becomes (0, 4).
|f(x)| = x² − 4 for x ≤ −2 or x ≥ 2
The outside portions stay unchanged. Range: y ≥ 0; corners occur at (±2, 0).
03 / Sketch y = f(|x|)
When x ≥ 0, |x| = x, so keep the original right-hand part. When x < 0, |x| = −x: the new output comes from the corresponding positive input. Mirror the right-hand part across the y-axis.
The old negative-input part is not used. Do not reflect it onto the right: that would mix two different rules. The resulting graph is symmetric about the y-axis wherever it is defined.
If (2, −3) lies on f, then (2, −3) and (−2, −3) lie on f(|x|)
An output can remain negative.
A point (−2, 5) on f contributes nothing by itself
The value at new input −2 comes from original input +2.
The point at x = 0, when defined, stays once
There are not two distinct copies of a point on the mirror line.
04 / Compare the two
For f(x) = 2x − 2, output modulus gives |2x − 2|, with vertex (1, 0). Input modulus gives 2|x| − 2, with vertex (0, −2). They have different zeros, vertices and ranges.
Dashed: original f(x) = 2x − 2. Gold ring: source point used. Green point: transformed output.
x = −1: f(x) = −4, so |f(x)| = 4.
Keep the input; make the output nonnegative.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Domains and ranges
The domain of f(|x|) is {x: |x| belongs to the domain of f}. Its range comes only from original inputs that are nonnegative and allowed. Negative-input outputs of f may disappear entirely.
f(x) = x − 1, defined only for x ≥ 2
The original range is y ≥ 1.
|f(x)| = x − 1, still only for x ≥ 2
All original outputs were already positive.
f(|x|) = |x| − 1, for x ≤ −2 or x ≥ 2
The middle interval (−2, 2) is excluded. Range remains y ≥ 1.
If f has no allowed nonnegative inputs, f(|x|) has an empty real domain. A formula sketch alone cannot supply missing allowed inputs.
06 / Other graph families
f(x) = 1/(x − 1)
|f(x)| keeps domain x ≠ 1 and has positive branches. f(|x|) = 1/(|x| − 1) has domain x ≠ ±1, with vertical asymptotes at both −1 and 1. It is negative for |x| < 1.
f(x) = eˣ − 2
|f(x)| has range [0, ∞). f(|x|) = e^|x| − 2 has minimum −1 at x = 0, so its range is [−1, ∞).
f(x) = sin x, using radians
|sin x| is nonnegative. sin(|x|) is even but still takes negative values, such as at x = ±3π/2. Both are defined for all real inputs.
An even function f, such as x² − 4
f(|x|) = f(x) when the domain is symmetric: the original graph already matches its right-hand mirror.
07 / Work from a piecewise graph
Suppose f(x) = x + 3 for −2 ≤ x < 0 and f(x) = x − 1 for 0 ≤ x ≤ 3. Its right-hand part alone determines f(|x|): the new rule is |x| − 1 for −3 ≤ x ≤ 3. The old values from negative inputs are discarded.
For |f(x)|, both original branches remain in use. The left branch x + 3 is positive throughout its domain, so it stays. On the right, x − 1 is reflected only for 0 ≤ x < 1. The output-modulus domain remains [−2, 3].
Keep hollow and filled endpoints consistent with the relevant branch restrictions. Reflecting coordinates does not turn an excluded endpoint into an included one.
08 / Your turn
Give exact coordinates and any domain exclusions.
The point (−3, −5) lies on f. What point is guaranteed on |f(x)|? Does it determine the value of f(|x|) at x = −3?
Output modulus keeps the input; input modulus uses f(3).
(−3, 5) lies on |f(x)|. The given point does not determine f(|−3|) = f(3).
If (4, −2) lies on f, which points must lie on f(|x|)?
Both inputs ±4 are sent to original input 4.
(4, −2) and (−4, −2). The output remains negative.
For f(x) = 3x − 6, state the vertices and ranges of |f(x)| and f(|x|).
Compare |3x − 6| with 3|x| − 6.
|f(x)| has vertex (2, 0) and range [0, ∞). f(|x|) has vertex (0, −6) and range [−6, ∞). Both domains are all real numbers.
For y = |x² − 9|, find the zeros, the point at x = 0 and the range.
Reflect the original graph between its roots.
The zeros are x = ±3. At x = 0 the output is 9. The range is [0, ∞); the central portion is 9 − x² for −3 ≤ x ≤ 3.
f(x) = x + 2 for 1 < x ≤ 4. Give the domain and range of f(|x|).
Solve 1 < |x| ≤ 4.
Domain: [−4, −1) ∪ (1, 4]. The outputs are |x| + 2, so the range is (3, 6]. Both inner endpoints are excluded.
For f(x) = 1/(x − 2), find the excluded inputs of |f(x)| and f(|x|).
Taking output modulus does not change the denominator. Input modulus does.
|f(x)| excludes x = 2. f(|x|) excludes x = ±2 because |x| − 2 = 0 there. These are their respective vertical asymptotes.
Evaluate |sin x| and sin(|x|) at x = −3π/2.
First evaluate the relevant sine input in each expression.
sin(−3π/2) = 1, so |sin x| = 1. But sin(|−3π/2|) = sin(3π/2) = −1.
Find the range and zeros of e^|x| − 3.
Since |x| ≥ 0, e^|x| ≥ 1.
The minimum is −2 at x = 0, giving range [−2, ∞). The zeros satisfy |x| = ln 3, so x = ±ln 3.
09 / Recap
Section 1 of 9 · Two different operations