Map the whole interval, not only its endpoints · source reference

Complete question

(a) $f(x)=x^2+\lambda x+2$ for real $x\le2$, with constant lambda. (i) Find the values of lambda for which an inverse exists. [2] (ii) With $\lambda=-4$, define the inverse, including its domain. [3] (b) $g(x)=x^2+x+1$ for all real x, and $h(x)=x+4/x$ for real $x>0$. (i) Show hg exists. [2] (ii) Find its range. [2]

[modified 2011/HCI/Prelim/II/2]

Printed working

Part (a)(i)

For $y=x^2+\lambda x+2$, $dy/dx=2x+\lambda=0$ gives $x=-\lambda/2$, the x-coordinate of the minimum.

For $f^{-1}$ to exist, $f$ must be one-one. Thus $-\lambda/2\ge2\Rightarrow\lambda\le-4$.

Alternative method to find the turning-point abscissa:

$$f(x)=x^2+\lambda x+2=\left(x+\frac\lambda2\right)^2-\frac{\lambda^2}4+2.$$

Thus $x=-\lambda/2$ is again the minimum abscissa. The two source diagrams compare a minimum to the right of 2 (injective restricted branch) and a minimum to the left of 2 (horizontal line $y=2$ cuts the restricted graph twice). Their shaded domain is $x\le2$.

Part (a)(ii)

Let $y=x^2-4x+2$, $x\le2$. Then $y=(x-2)^2-2$, giving $x=2\pm\sqrt{y+2}$.

Since $x\le2$, reject $x=2+\sqrt{y+2}$ and take $x=2-\sqrt{y+2}$. $D_{f^{-1}}=R_f=[-2,\infty)$.

Therefore $f^{-1}:x\mapsto2-\sqrt{x+2}$, $x\ge-2$.

Part (b)(i)

$$g(x)=x^2+x+1=\left(x+\frac12\right)^2+\frac34,\qquad R_g=[3/4,\infty).$$

The graph labels minimum $(-1/2,3/4)$. Since $R_g=[3/4,\infty)\subseteq D_h=(0,\infty)$, $hg$ exists (shown).

Part (b)(ii)

Use $R_g=[3/4,\infty)$ as the new domain of $h(x)=x+4/x$. The source graph labels its minimum $(2,4)$, vertical asymptote $x=0$, oblique asymptote $y=x$, the domain ray beginning at $3/4$ and a vertical range ray beginning at 4.

$$D_g=\mathbb R\xrightarrow{g}R_g=[3/4,\infty)\xrightarrow{h}R_{hg}=[4,\infty).$$

Thus $R_{hg}=[4,\infty)$.

Teacher margin notes

For $f$ to be1–1, the minimum at $x=-\lambda/2$ must be to the right of $x=2$, i.e. $-\lambda/2\ge2$. If it is to the left of $x=2$, as in the second diagram, $f$ is not1–1 because horizontal line $y=2$ cuts its graph at two points.

To find $f^{-1}(x)$: (1) let $y=f(x)$ so $x=f^{-1}(y)$; (2) make $x$ the subject and choose the expression corresponding to the domain of $f$; (3) replace $y$ with $x$; (4) define $f^{-1}$ — both rule and domain. Note $D_{f^{-1}}=R_f$.

Given two functions $g,h$, $hg$ is $g$ (first function) followed by $h$ (second function). For $hg$ to exist, range of $g\subseteq$ domain of $h$.

To find $R_{hg}$: (1) find $R_g$ using the graph of $g$; (2) sketch the graph of the second function $h$; (3) range of the first function becomes the restricted domain of $h$; (4) use that new domain and the graph of $h$ to find its range, which is $R_{hg}$.

Source diagrams

Provenance

Question page references: 2. Solution page references: 8, 9. 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.