When can two functions compose? · source reference

Complete question

The functions are $f(x)=\sqrt{x-2}$ for $x\ge2$, and $g(x)=3x^2-6x+5$ for $0\le x\le4$, with real inputs. (i) Sketch y=f(x) and state its range. [2] (ii) Find the inverse of f and its domain. [2] (iii) Show that fg exists. [2] (iv) Find fg(x) and its range. [3] (v) With g now defined on all real numbers, describe fully a sequence of transformations taking y=g(x) to $y=x^2-1$. [3]

[2023/MI/Promo/PU2/I/8]

Printed working

Part (i)

From the square-root graph beginning at the closed point $(2,0)$, $R_f=[0,\infty)$.

Part (ii)

$$y=\sqrt{x-2}\Rightarrow y^2=x-2\Rightarrow x=y^2+2.$$

The printed line says “where $x\ge0$ (Note: $x=f^{-1}(y)$)”. The inverse input in this rearrangement is $y\ge0$; the final domain below supplies that restriction.

$$f^{-1}(x)=x^2+2,\qquad D_{f^{-1}}=R_f=[0,\infty).$$

Part (iii)

$$R_g=[2,29],\qquad D_f=[2,\infty),\qquad R_g\subseteq D_f.$$

Thus $fg$ exists. The graph of $g$ labels $(0,5)$, minimum $(1,2)$ and closed endpoint $(4,29)$. The original nested interval diagram places the bounded $R_g$ interval inside the rightward $D_f$ ray, with labels 2 and 29.

Part (iv)

$$\begin{aligned} fg(x)&=f[g(x)]\\ &=f(3x^2-6x+5)\\ &=\sqrt{(3x^2-6x+5)-2}\\ &=\sqrt{3x^2-6x+3}. \end{aligned}$$

Method 1: $[0,4]\xrightarrow{g}[2,29]\xrightarrow{f}?$, with domains/ranges labelled $D_g$, $R_g$, $R_{fg}$. Sketch $y=f(x)=\sqrt{x-2}$ for $2\le x\le29$. The graph labels closed points $(2,0)$ and $(29,\sqrt{27})$ and the level $\sqrt{27}$. Therefore $R_{fg}=[0,\sqrt{27}]$.

Method 2: Sketch $y=fg(x)=\sqrt{3x^2-6x+3}$, $0\le x\le4$. The V has vertex $(1,0)$ and closed endpoint $(4,\sqrt{27})$; a vertical double-headed arrow labels $R_{fg}$. Thus $R_{fg}=[0,\sqrt{27}]$ again.

Part (v): three printed transformation methods

Let $y=g(x)=3x^2-6x+5=3(x-1)^2+2$.

Method 1:

1. Replace $x$ by $x+1$: $y=3((x+1)-1)^2+2=3x^2+2$. Translate in the negative x-direction by 1 unit.
2. Replace $y$ by $y+5$: $y+5=3x^2+2$, so $y=3x^2-3$. Translate in the negative y-direction by 5 units.
3. Replace $y$ by $3y$: $3y=3x^2-3\Rightarrow y=x^2-1$. Scale parallel to the y-axis by scale factor $1/3$.

Method 2:

1. Replace $x$ by $x+1$: $y=3((x+1)-1)^2+2=3x^2+2$. Translate in the negative x-direction by 1 unit.
2. Replace $y$ by $3y$: $3y=3x^2+2\Rightarrow y=x^2+2/3$. Scale parallel to the y-axis by scale factor $1/3$.
3. Replace $y$ by $y+5/3$: $y+5/3=x^2+2/3\Rightarrow y=x^2-1$. Translate in the negative y-direction by $5/3$ units.

Method 3:

1. Replace $x$ by $x+1$: $y=3((x+1)-1)^2+2=3x^2+2$. Translate in the negative x-direction by 1 unit.
2. Replace $x$ by $x/\sqrt3$: $y=3(x/\sqrt3)^2+2=x^2+2$. Scale parallel to the x-axis by scale factor $\sqrt3$.
3. Replace $y$ by $y+3$: $y+3=x^2+2\Rightarrow y=x^2-1$. Translate in the negative y-direction by 3 units.

Teacher margin notes

To find the range of $f$, sketch the graph of $f$ and, from the graph, find the range of y-values.

$y=f(x)\Leftrightarrow x=f^{-1}(y)$. Make $x$ the subject where $x=f^{-1}(y)$. Then replace $y$ by $x$ to find $f^{-1}(x)$.

Note that the range of $g$ is from the y-value of the minimum point $(1,2)$ to the y-value of the endpoint $(4,29)$. Do not just consider the y-values of the two endpoints to find the range of $g$.

Treat $R_g$ as the new domain for the graph of $f$ to find $R_{fg}$.

Source diagrams

Review qualification

Endpoint checks alone are insufficient: the minimum of g occurs inside its domain.

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

Provenance

Question page references: 1. Solution page references: 1, 2, 3. 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.