D11 - Modulus and reciprocal graphs

Reciprocal graph

The graph of $y=1/f(x)$ consists of the points $(x,1/f(x))$ for which $f(x)$ is defined and nonzero. It is a reciprocal of the output, not an inverse function: $1/f(x)$ and $f^{-1}(x)$ mean different operations.

Worked example 1: two transformations of a line

Question. Sketch the key features of $y=|f(x)|$ and $y=1/f(x)$ for $f(x)=x-1$.

Plan. Transform useful output values and examine what happens near the original zero $x=1$.

The modulus graph is a V with vertex $(1,0)$, left branch $y=1-x$ and right branch $y=x-1$. Its domain is all real numbers.

For the reciprocal, approaching 1 from the left gives a small negative denominator, so the output tends to $-\infty$; approaching from the right gives $+\infty$. Far from 1, the denominator grows in magnitude, so the output approaches zero.

Answer. The modulus graph has vertex $(1,0)$ and vertical-axis intercept $(0,1)$. The reciprocal graph $y=1/(x-1)$ has asymptotes $x=1$ and $y=0$, vertical-axis intercept $(0,-1)$, a negative left branch and a positive right branch.

Check. The identities $|x-1|\ge0$ and $(x-1)\bigl(1/(x-1)\bigr)=1$ hold on their respective domains. In particular, at $x=3$, the transformed outputs are 2 and $1/2$, answering the opening question for this function.

Worked example 2: scaling before taking a reciprocal

Question. Sketch the key features of $y=|g(x)|$ and $y=1/g(x)$ for $g(x)=2(x-1)$.

Plan. Use the previous graphs and track how doubling each original output changes the new ones.

The modulus graph keeps vertex $(1,0)$, with gradients $-2$ and 2, and passes through $(0,2)$. The reciprocal still has asymptotes $x=1$ and $y=0$, but now passes through $(0,-1/2)$ and $(2,1/2)$. Its branches keep the same signs.

Answer. A steeper V with unchanged vertex, and a reciprocal curve whose finite outputs are half those in Worked example 1. The reciprocal domain is $x\ne1$.

Check. At $x=3/2$, $g(x)=1$, so both transformed outputs equal 1. At $x=1/2$, $g(x)=-1$, so the modulus output is 1 while the reciprocal output is $-1$. Both match the formulas above.

Watch out

Wrong: reciprocal graphs reflect in the line $y=x$. That describes exchanging input and output when an inverse function exists. A reciprocal keeps $x$ fixed and changes $u$ to $1/u$.

Key idea

A reciprocal graph has no zero: the reciprocal of a finite nonzero real number cannot equal zero. It may approach zero, but a missing original input cannot be filled merely because a simplified formula has a finite value there.

Worked example 3: an original asymptote becomes a hole

Question. Let $f(x)=\frac{x+1}{x-2}$, with $x\ne2$. Find the zero and vertical asymptote of $|f(x)|$. For $h(x)=1/f(x)$, find the domain, asymptotes, hole and points fixed by reciprocation.

Plan. Preserve the original domain, exclude the original zero, and then simplify the reciprocal on that domain.

The modulus has zero $x=-1$ and vertical asymptote $x=2$; reflecting a branch does not bound its magnitude. It also approaches level 1 at both far ends.

For $h$, the original zero becomes the vertical asymptote $x=-1$. The rewritten form shows horizontal asymptote $y=1$. As $x$ approaches 2, the simplified formula approaches 0, but the original $f$ is undefined at 2, so $h$ has a hole at $(2,0)$, not a zero.

To find the fixed points, solve the original level equations:

Answer. For $|f|$: zero $-1$, vertical asymptote $x=2$. For $h$: domain $\mathbb R\setminus\{-1,2\}$, asymptotes $x=-1$ and $y=1$, hole $(2,0)$, and one fixed point $(1/2,-1)$.

Check. $f(1/2)=-1$, whose reciprocal is $-1$. Also $f(0)=-1/2$ and $h(0)=-2$, so their product is 1; neither graph's horizontal asymptote is zero.

Worked example 4: fixed levels after scaling

Question. Let $g(x)=\frac{2(x+1)}{x-2}$, with $x\ne2$. Find the corresponding features for $|g(x)|$ and $k(x)=1/g(x)$.

Plan. Retain the original exclusion and zero, then solve $g(x)=\pm1$ for the fixed points.

The modulus still vanishes at $-1$ and has vertical asymptote $x=2$; its horizontal limiting level is now 2. The reciprocal still excludes $-1$ and 2, has vertical asymptote $x=-1$, and approaches zero at the hole $(2,0)$. Its far-end level is now $1/2$.

Answer. For $|g|$: zero $-1$, vertical asymptote $x=2$. For $k$: domain $\mathbb R\setminus\{-1,2\}$, asymptotes $x=-1$ and $y=1/2$, hole $(2,0)$, and fixed points $(-4,1)$ and $(0,-1)$.

Check. $g(-4)=1$ and $g(0)=-1$, so reciprocation leaves both points fixed. The entire formula $k(x)=h(x)/2$ also confirms the halved horizontal level and unchanged excluded inputs.

Watch out

Wrong: every reciprocal graph has horizontal asymptote $y=0$. That is true when the original magnitude grows without bound at the relevant far end. If the original tends to a nonzero constant $L$, the reciprocal tends to $1/L$.