Distinguish input and output modulus

Keep output modulus and input modulus distinct

First locate the modulus bars. In $|f(x)|$, they act on the output; in $f(|x|)$, they act on the input. These rules retain different parts of the original graph.

Output modulus: keep the upper part

For $y=|f(x)|$:

1. Keep every portion with $y\ge0$.
2. Reflect each portion with $y<0$ in the x-axis.

The outputs become non-negative, the zeros stay in place, and the original domain is preserved. The quadratic activity folds the lower arc of $x^2-1$ upwards. Example 13 keeps the cubic’s zeros $-1,2,3$ and reflects its negative arcs.

In Example 14, the rational graph keeps its x-intercept $a/2$ and y-intercept 1. Its horizontal asymptote changes from $-2$ to 2, while $x=-a$ remains excluded. Self-Practice 5 gives $|x+4|$, with vertex $(-4,0)$; its sketch is independently derived.

Input modulus: keep the right-hand part

For $y=f(|x|)$:

1. Keep only the original portion with $x\ge0$.
2. Reflect that retained portion in the y-axis to create the left-hand part. Discard the original left-hand part.

For $x\ge0$, $|x|=x$, so use $f(x)$. For $x<0$, $|x|=-x$, so use $f(-x)$, wherever defined. The result has y-axis symmetry, but outputs can still be negative. Its domain consists of inputs x for which $|x|$ belongs to the original domain.

The logarithm activity keeps $(0,\ln2)$ and mirrors the right-hand part. The resulting $\ln(|x|+2)$ is defined for all real x and has no vertical asymptote, unlike $\ln(x+2)$.

In Example 15, discard the original root $-1$. Keep roots 2 and 3, and mirror them to $-2,-3$. The negative troughs stay below the x-axis; the sketch includes all four printed approximate extremum labels. Self-Practice 6 gives $|x|+4$, with vertex $(0,4)$ and no x-intercept. Its independently derived sketch differs from Self-Practice 5.

Read a schematic graph carefully

Example 16 supplies a sketch, not a formula. Mirror its positive-input features: poles at $\pm1$, roots at $\pm3$, horizontal asymptote $y=1$ and central point $(0,3)$. These features do not determine a unique formula.

The centre is a corner local minimum, although it is not a differentiable stationary point. The Teacher’s phrase “is not a turning point” uses a narrower meaning; it should not be read as denying the minimum visible in the sketch. The original wording remains in the source review and native text.

Check before the examples: a point $(2,-3)$ becomes $(2,3)$ under output modulus. Under input modulus it stays at $(2,-3)$ and is copied to $(-2,-3)$. Input modulus does not make the output positive.

Source correction: the logarithm activity’s later “Hence to sketch” line omits the modulus bars. Use the displayed piecewise rule above; the original typo remains in the native text.