Functions · edition and printed answer key

Algebra & functions

Calculus

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PorismaFunctions

R_f=[0,\infty)

Question	Key as printed
1	(i) $R_f=[0,\infty)$ ; (ii) $f^{-1}(x)=x^2+2$ , domain $[0,\infty)$ ; (iv) $fg(x)=\sqrt{3x^2-6x+3}$ , $R_{fg}=[0,\sqrt{27}]$
2	(a)(i) $D_{f^{-1}}=R_f=(-a,e^7-a]$ ; (a)(ii) $a=-34$ ; (b)(i) $h(9)=3/2$
3	(i) $(0,2]$ ; (iii) $1/5$ , $f^{-1}(x)=(1+\sqrt{2/x-1})/5$ ; (iv) inverse-reflection diagram
4	(a)(i) $\lambda\le-4$ ; (a)(ii) $f^{-1}:x\mapsto2-\sqrt{x+2}$ , $x\ge-2$ ; (b)(ii) $[4,\infty)$
5	(a) exists; (b)(i) $gh:x\mapsto\ln(x^2-2x+2)$ , $x>0$ , $R_{gh}=[0,\infty)$ ; (b)(ii) log/inverse/identity diagram; (b)(iii) $[1,\infty)$
6	(i) $f^{-1}:x\mapsto1/(1-x)$ , $x\in\mathbb R$ , $x\ne0,1$ ; (iii) $f^3(x)=x$ ; (iv) $g(x)=e^x$
7	(i) $1/2$ ; (ii) $f^{-1}:x\mapsto-1/2+1/(1-x)$ , $0\le x<1$ ; (iii) $R_{gf^{-1}}=[\pi/4,\pi/2)$
8	(ii) $1-x^2/2+x^3/2+\cdots$ ; (iii) $h^{-1}(-1)=1$ and the piecewise rule $e^{-x}-1$ for $x>-\ln2$ , $1$ for $x=-1$
9	(i) $2\sin(x+\pi/3)$ ; (ii) largest $k=7\pi/6$ ; printed inverse $\sin^{-1}(x/2)-\pi/3$ , $-2\le x\le2$ ; comparison interval $[\pi/6,2]$ . The inverse branch is corrected separately in the question record.

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