Restrict the domain to choose an inverse · source reference

Complete question

For all real x, $f(x)=2/[1+(5x-1)^2]$. (i) Find its range. [2] (ii) Explain why it has no inverse. [1] (iii) Restrict its domain to $x\ge k$: find the least k that gives an inverse, and find that inverse. [3] (iv) Sketch the restricted f and its inverse together, clearly showing their relationship. [2] (v) Show algebraically that a solution of $f(x)=f^{-1}(x)$ satisfies $25x^3-10x^2+2x-2=0$. [2]

[2011/IJC/Prelim/I/9]

Printed working

Part (i)

The graph is symmetric about $x=1/5$, with maximum $(1/5,2)$ and horizontal asymptote $y=0$. The marked line $y=1$ intersects the unrestricted graph twice. $R_f=(0,2]$.

Part (ii): graphical method

The horizontal line $y=1$ cuts the graph of $f$ at two points ($>1$), hence $f$ is not one-one and does not have an inverse.

Part (ii): counter-example — non-graphical

$$f(0)=\frac2{1+(0-1)^2}=1,$$ $$f\left(\frac25\right)=\frac2{1+(5(2/5)-1)^2}=\frac2{1+(2-1)^2}=1.$$

Since $f(0)=f(2/5)=1$ but $0\ne2/5$ and both inputs lie in $\mathbb R$, $f$ is not one-one and does not have an inverse.

Part (iii)

Least $k=1/5$. Let $y=f(x)=2/[1+(5x-1)^2]$, $x\ge1/5$:

$$1+(5x-1)^2=\frac2y\Rightarrow5x-1=\pm\sqrt{\frac2y-1}.$$

Since $x\ge1/5$, $5x-1\ge0$. Hence

$$x=\frac15\left(1+\sqrt{\frac2y-1}\right) \Rightarrow f^{-1}(x)=\frac15\left(1+\sqrt{\frac2x-1}\right),\qquad0<x\le2.$$

Part (iv)

The restricted graph and its inverse are reflected about $y=x$, with endpoints $(1/5,2)$ and $(2,1/5)$ respectively. Both axes use the same scale. The inverse has vertical asymptote $x=0$, while the restricted function tends to $y=0$.

Part (v): printed derivation

$$f(x)=f^{-1}(x)\Rightarrow f(x)=x\Rightarrow\frac2{1+(5x-1)^2}=x.$$

Therefore $x(1+25x^2-10x+1)=2$, giving $25x^3-10x^2+2x-2=0$ (shown).

The implication to the fixed-point equation is justified for this restricted curve; the authored article supplies the stronger domain check that excludes extra unrestricted two-cycle roots.

Teacher margin notes

The range of $f$ is the set of y-values that $f$ can take. Sketch the graph of $f$ according to the given domain before determining the range.

Graphical method: use the graph of $f$ sketched in (i). To show $f$ is not1–1, give a counter-example of one particular horizontal line that cuts the graph of $f$ more than once.

Counter-example: give two different values of x that result in the same image y.

Deduce the least $k$ from the graph of $f$. State the domain of $f$. To find $f^{-1}(x)$: (1) let $y=f(x)\Rightarrow x=f^{-1}(y)$; (2) make $x$ the subject, and if there are two possible expressions, choose the one corresponding to values $x$ can take in the domain of $f$; (3) replace $y$ with $x$. Note: $f^{-1}(x)\ne1/f(x)$.

When sketching $f$ and $f^{-1}$ together, use the same scale for x- and y-axes; their graphs must be symmetrical about $y=x$. Always include endpoints. In this case the two curves intersect only on $y=x$, so the two curves and the line meet at the same point.

To solve $f(x)=f^{-1}(x)$, it may be easier to solve $f(x)=x$ or $f^{-1}(x)=x$.

Source diagrams

Review qualification

The full printed solution sketch labels a horizontal asymptote $y=1$, whereas the question short key and the formula imply $y=0$. The native graph uses the mathematically correct asymptote $y=0$; $y=1$ is the separate horizontal-line test level. This preserves the printed discrepancy explicitly.

The general claim that f and its inverse intersect only on y=x is false for decreasing functions. The working here includes the extra admissibility check.

The corrected or qualified authored derivation is retained in the question article.

Provenance

Question page references: 1, 2. Solution page references: 5, 6, 7. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.