Invert, then repeat periodically · source reference

Complete question

(a) $f(x)=e^{x+3}-a$ for real $x\le4$; $g(x)=2x+5$ for all real x. The parameter a is constant. (i) Find the inverse of f and its domain in terms of a. [3] (ii) Given that the composite exists and $g^{-1}f(-3)=15$, find a without finding the composite rule. [2] (b) $h(x)=2^x$ for $0<x\le2$, and $h(x)=6-x^2/2$ for $2<x\le3$, with $h(x)=h(x-3)$ for all real x. (i) Find h(9). [1] (ii) Sketch h on $[-1,4]$. [4]

[2023/ACJC/Promo/5]

Printed working

Part (a)(i)

Let $y=f(x)$:

$$y=e^{x+3}-a\Rightarrow e^{x+3}=y+a\Rightarrow x=\ln(y+a)-3.$$

Thus $f^{-1}(x)=\ln(x+a)-3$ and $D_{f^{-1}}=R_f=(-a,e^7-a]$.

Part (a)(ii)

$$g^{-1}f(-3)=15\Rightarrow f(-3)=g(15)\Rightarrow e^{-3+3}-a=2(15)+5.$$

Therefore $1-a=35\Rightarrow a=-34$.

Part (b)(i)

$$h(9)=h(6+3)=h(6)=h(3+3)=h(3)=6-\frac12(3)^2=\frac32.$$

Part (b)(ii)

The periodic sketch has period 3 and shows the following branches on $-1\le x\le4$:

Branch	Domain shown	Endpoint conventions retained
$6-\tfrac12(x+3)^2$	$[-1,0]$	closed $(-1,4)$ and $(0,1.5)$
$2^x$	$(0,2]$	open $(0,1)$, closed $(2,4)$
$6-\tfrac12x^2$	$(2,3]$	joins closed $(2,4)$, closed $(3,1.5)$
$2^{x-3}$	$(3,4]$	open $(3,1)$, closed $(4,2)$

The source labels O and every endpoint above; the period-boundary values at 0 and 3 come from the right-closed quadratic piece, not the open exponential piece.

Teacher margin notes

Make $x$ the subject.

Source diagrams

Provenance

Question page references: 1. Solution page references: 4. 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.