1.2 - Modulus graphs and inequalities

Output modulus

The graph $y=|f(x)|$ keeps the parts of $y=f(x)$ on or above the $x$-axis and reflects the parts below it in that axis. The domain is unchanged.

Worked example 1: reflecting one negative interval

Question. Describe the graph of $y=|(x-2)/(x+1)|$.

Plan. Use the zero and pole to identify the negative interval, then reflect it.

The original function is positive for $x<-1$, negative for $-1<x<2$, and positive for $x>2$. Therefore

The zero remains $(2,0)$, and $(0,-2)$ becomes $(0,2)$. At $-1$, both branches now tend to $+\infty$. The horizontal asymptote remains $y=1$. Since the original derivative at $2$ is $3/(2+1)^2=1/3$, reflection reverses the left gradient, creating a corner.

Answer. Reflect only the interval $-1<x<2$. The resulting graph is nonnegative, has a corner at $(2,0)$, asymptotes $x=-1$, $y=1$, and intercept $(0,2)$.

Check. At $x=1$, the original height is $-1/2$, so the new one is $1/2$. At $x=-2$, the original height is $4$, so it stays $4$.

Worked example 2: a repeated zero stays smooth

Question. Which part of $g(x)=(x-2)^2/(x+1)$ is reflected in $y=|g(x)|$? Does a corner appear at $x=2$?

Plan. Use the nonnegative squared numerator to find signs, then inspect the curve near its zero.

For $x<-1$, the denominator is negative and the numerator is positive, so $g<0$. For $x>-1$, $g\ge0$. Thus reflect the left branch $x<-1$ only. Near $2$, the denominator is positive on both sides, so the graph is unchanged there.

More explicitly,

Answer. Reflect $x<-1$; retain $x>-1$. The point $(2,0)$ is a smooth minimum, with no corner.

Check. At $x=2+h$, $g=h^2/(3+h)$, which is nonnegative for sufficiently small positive or negative $h$. Taking its modulus makes no local change.

Watch out

Modulus does not automatically create a corner at every zero. Check whether the original graph crosses or merely touches, and whether its gradient is nonzero there.

Input modulus

To sketch $y=f(|x|)$, keep the original graph for $x\ge0$, then reflect that retained graph in the $y$-axis. The result is even: $f(|-x|)=f(|x|)$.

Its domain consists of inputs for which $|x|$ belongs to the original domain.

Worked example 3: a negative pole disappears

Question. Describe $y=f(|x|)$, where $f(x)=(x-2)/(x+1)$.

Plan. Keep the original graph for nonnegative inputs and reflect it in the $y$-axis.

The denominator is at least $1$, so the domain is all real inputs and there are no vertical asymptotes. The zeros satisfy $|x|=2$, giving $x=\pm2$. At $0$, the output is $-2$. As $x\to\pm\infty$, $|x|\to\infty$ and the output tends to $1$ from below.

Answer. An even graph with zeros $(\pm2,0)$, minimum $(0,-2)$, horizontal asymptote $y=1$, and no vertical asymptote. There is a corner at $(0,-2)$, since the reflected one-sided gradients are $-3$ and $3$.

Check. Both $x=3$ and $x=-3$ give $1/4$. The original exclusion $-1$ does not survive: at new input $-1$, the function uses original input $1$, giving $-1/2$.

Worked example 4: a positive pole is copied

Question. Describe $y=g(|x|)$, where $g(x)=(x-2)/(x-1)$.

Plan. Apply the original restriction to $|x|$, then copy the right-hand features.

Thus $x\ne\pm1$. The original positive zero $2$ gives zeros $\pm2$, and the original $y$-intercept is $(0,2)$. The horizontal asymptote is $y=1$ at both ends. At $x=1$, the limits are $+\infty$ from the left and $-\infty$ from the right; symmetry determines those at $-1$.

Answer. An even graph with vertical asymptotes $x=\pm1$, horizontal asymptote $y=1$, zeros $(\pm2,0)$, and intercept $(0,2)$.

Check. At $x=\pm1/2$, the output is $3$; at $x=\pm3/2$, it is $-1$. These agree with a positive central region and reflected negative regions just outside the poles.

Worked example 5: one finite crossing

Question. Solve $|(x-2)/(x+1)|\le1$.

Plan. Find where the modulus curve meets $y=1$, then use an equivalent squared inequality to confirm the interval.

For $x\ne-1$,

The boundary has original height $-1$. The original height $+1$ is never attained, so there is only one finite crossing. The excluded input $-1$ is outside the proposed interval.

Answer. $x\ge1/2$.

Check. At $x=1/2$, the modulus is $1$; at $x=1$, it is $1/2$; at $x=0$, it is $2$. At large positive inputs the curve stays below $1$, consistently approaching its asymptote.

Worked example 6: two crossings and a pole between them

Question. Solve $|(x-2)/(x+1)|\le2$.

Plan. Find the two crossing inputs and use the sign of the squared comparison.

For $x\ne-1$,

The product is nonnegative for $x\le-4$ or $x\ge0$. At $-4$, the original height is $2$; at $0$, it is $-2$. The modulus graph lies above height $2$ between those crossings, with its pole inside that gap.

Answer. $x\le-4$ or $x\ge0$.

Check. At $x=-5$, the modulus is $7/4<2$; at $x=-2$, it is $4>2$; at $x=1$, it is $1/2<2$. Equality holds at both stated boundaries.

Watch out

Strict inequalities exclude crossing points; poles are excluded in both strict and non-strict inequalities. A hole inside a selected interval must also be removed.