Solution. An arrow diagram puts the first components in a left-hand oval and the second components in a right-hand one, with an arrow from each first component to its partner.

The test is always the same: a set of ordered pairs is a function when every first component has exactly one arrow leaving it.

(i). The first components are \(0\), \(2\), \(1\), \(3\) — all different, so each has exactly one arrow. It is a function.

(ii). The first components are \(1\), \(3\), \(2\) — all different. It is a function.

(iii). The first component \(1\) appears three times, sending \(1\) to \(a\), to \(b\) and to \(d\). It is not a function, because a function must give one output for each input.

(iv). The first components are \(1\), \(2\), \(3\), \(4\) — all different. It is a function, even though three of them share the output \(b\).

Solution. The domain is the set of first components and the range the set of second components. The inverse is formed by reversing every pair.

(a). \[\text {domain of } f=\{1,2,5\},\qquad \text {range of } f=\{5,9,21\}.\] Reversing each pair, \[f^{-1}=\{(5,1),(9,2),(21,5)\},\] \[\text {domain of } f^{-1}=\{5,9,21\},\qquad \text {range of } f^{-1}=\{1,2,5\}.\]

(b). \[\text {domain of } f=\{-2,-1,0,2\},\qquad \text {range of } f=\{-8,-1,0,8\}.\] \[f^{-1}=\{(0,0),(8,2),(-1,-1),(-8,-2)\},\] \[\text {domain of } f^{-1}=\{-8,-1,0,8\},\qquad \text {range of } f^{-1}=\{-2,-1,0,2\}.\]

Solution. With a piecewise function, the first job is always to decide which line applies, and only then to substitute.

(a). \[f\left (\tfrac {1}{2}\right ):\quad \tfrac {1}{2}\geq 0,\text { so use } 3x+1:\qquad 3\left (\tfrac {1}{2}\right )+1=\tfrac {3}{2}+1=\tfrac {5}{2}.\] \[f(0):\quad 0\geq 0,\text { so use } 3x+1:\qquad 3(0)+1=1.\] \[f(-3):\quad -3<0,\text { so use } 2x-3:\qquad 2(-3)-3=-6-3=-9.\]

(b). \[f(-5):\quad -5<0,\text { so } f(-5)=-5.\] \[f\left (\tfrac {5}{2}\right ):\quad 0\leq \tfrac {5}{2}\leq 3,\text { so use } 2x^2+3: \qquad 2\left (\tfrac {25}{4}\right )+3=\tfrac {25}{2}+3=\tfrac {31}{2}.\] \[f(0):\quad 0\leq 0\leq 3,\text { so use } 2x^2+3:\qquad 2(0)+3=3.\] \[f(10):\quad 10>3,\text { so } f(10)=5.\]

Solution. (i). \[f(x+h)=3(x+h)-2=3x+3h-2,\] \[f(x+h)-f(x)=(3x+3h-2)-(3x-2)=3h,\] \[\frac {f(x+h)-f(x)}{h}=\frac {3h}{h}=3.\]

(ii). \[f(x+h)=2(x+h)^2-3(x+h)+1=2x^2+4xh+2h^2-3x-3h+1,\] \[f(x+h)-f(x)=4xh+2h^2-3h=h(4x+2h-3),\] \[\frac {f(x+h)-f(x)}{h}=4x+2h-3.\]

(iii). Combine the two fractions over a common denominator first: \[f(x+h)-f(x)=\frac {1}{2(x+h)}-\frac {1}{2x}=\frac {x-(x+h)}{2x(x+h)}=\frac {-h}{2x(x+h)}.\] Dividing by \(h\) means multiplying by \(\frac {1}{h}\): \[\frac {f(x+h)-f(x)}{h}=\frac {-h}{2x(x+h)}\times \frac {1}{h}=\frac {-1}{2x(x+h)}.\]

(iv). Again combine first: \[f(x+h)-f(x)=\frac {1}{\sqrt {x+h}}-\frac {1}{\sqrt {x}} =\frac {\sqrt {x}-\sqrt {x+h}}{\sqrt {x}\sqrt {x+h}}.\] The numerator is a difference of surds, so rationalise it by multiplying top and bottom by \(\sqrt {x}+\sqrt {x+h}\): \[\left (\sqrt {x}-\sqrt {x+h}\right )\left (\sqrt {x}+\sqrt {x+h}\right )=x-(x+h)=-h,\] \[f(x+h)-f(x)=\frac {-h}{\sqrt {x}\sqrt {x+h}\left (\sqrt {x}+\sqrt {x+h}\right )}.\] Now divide by \(h\): \[\frac {f(x+h)-f(x)}{h} =\frac {-1}{\sqrt {x}\sqrt {x+h}\left (\sqrt {x}+\sqrt {x+h}\right )}.\]

Solution. The domain is every \(x\) for which the formula makes sense. Two things break it: dividing by zero, and taking the square root of a negative number.

(i). A straight line, defined for every \(x\), and as \(x\) runs over all reals so does \(3x-7\): \[\text {domain}=\mathbb {R},\qquad \text {range}=\mathbb {R}.\]

(ii). The denominator vanishes when \[3x-5=0\quad \implies \quad x=\tfrac {5}{3}.\] \[\text {domain}=\mathbb {R}-\left \{\tfrac {5}{3}\right \}.\] For the range, no value of \(x\) makes the fraction zero, since a fraction is zero only when its numerator is, and the numerator is the constant \(2\). Every other value is attainable: \[\text {range}=\mathbb {R}-\{0\}.\]

(iii). A parabola, defined everywhere. Completing the square: \[x^2-3x=\left (x-\tfrac {3}{2}\right )^2-\tfrac {9}{4}.\] A square is never negative, so the smallest value is \(-\frac {9}{4}\), reached at \(x=\frac {3}{2}\): \[\text {domain}=\mathbb {R},\qquad \text {range}=\left [-\tfrac {9}{4},\ \infty \right ).\]

(iv). Here the root must be defined and non-zero, since it sits in a denominator. So we need \(2x-7>0\) rather than \(\geq 0\): \[2x-7>0\quad \implies \quad x>\tfrac {7}{2}.\] \[\text {domain}=\left (\tfrac {7}{2},\ \infty \right ).\] As \(x\) falls towards \(\frac {7}{2}\) the denominator shrinks to zero and \(f(x)\) grows without bound; as \(x\) grows the denominator grows and \(f(x)\) approaches zero without reaching it: \[\text {range}=(0,\ \infty ).\]

(v). A square root standing alone, so \(\geq 0\) suffices: \[3-x\geq 0\quad \implies \quad x\leq 3.\] \[\text {domain}=(-\infty ,\ 3],\qquad \text {range}=[0,\ \infty ),\] the range because a square root is never negative and takes every non-negative value as \(x\) runs down from \(3\).

Solution. \((f\circ g)(x)\) means \(f\left (g(x)\right )\): apply \(g\) first, then feed the result into \(f\).

(i). \[(f\circ g)(x)=f\left (x^2-3\right )=2\left (x^2-3\right )-6=2x^2-6-6=2x^2-12,\] \[(g\circ f)(x)=g(2x-6)=(2x-6)^2-3=4x^2-24x+36-3=4x^2-24x+33.\] Both \(f\) and \(g\) are defined for every real number, so both composites have domain \(\mathbb {R}\).

(ii). \[(f\circ g)(x)=f(3x-1)=\sqrt {(3x-1)-2}=\sqrt {3x-3}.\] The root needs \(3x-3\geq 0\), so \[\text {domain}=[1,\infty ).\] \[(g\circ f)(x)=g\left (\sqrt {x-2}\right )=3\sqrt {x-2}-1.\] Here \(g\) accepts anything, but \(f\) must be defined first, needing \(x-2\geq 0\): \[\text {domain}=[2,\infty ).\]

(iii). \[(f\circ g)(x)=f\left (\frac {2}{x}\right )=\frac {1}{\frac {2}{x}-1} =\frac {1}{\frac {2-x}{x}}=\frac {x}{2-x}.\] Two conditions apply. First \(g\) must be defined, so \(x\neq 0\). Second, \(f\) must accept \(g(x)\), which needs \(\frac {2}{x}\neq 1\), that is \(x\neq 2\): \[\text {domain}=\mathbb {R}-\{0,2\}.\] \[(g\circ f)(x)=g\left (\frac {1}{x-1}\right )=\frac {2}{\frac {1}{x-1}}=2(x-1)=2x-2.\] Here \(f\) must be defined, so \(x\neq 1\). The value \(f(x)=\frac {1}{x-1}\) is never zero, so \(g\) raises no further objection: \[\text {domain}=\mathbb {R}-\{1\}.\]

Solution. (a). There is no need to build the whole composite — substitute the number and work outwards. \[(f\circ g)(4)=f\left (g(4)\right )=f(4+4)=f(8)=8^2-2=62.\] \[(g\circ f)(4)=g\left (f(4)\right )=g\left (4^2-2\right )=g(14)=14+4=18.\]

(b). \[(f\circ g)(x)=f\left (x^2-4\right )=\sqrt {x^2-4}.\] For the domain, \(x^2-4\geq 0\), that is \(x^2\geq 4\), so \(x\leq -2\) or \(x\geq 2\): \[\text {domain}=(-\infty ,-2]\cup [2,\infty ).\] The inner expression \(x^2-4\) takes every value from \(0\) upwards on that domain, so \[\text {range}=[0,\infty ).\]

\[(g\circ f)(x)=g\left (\sqrt {x}\right )=\left (\sqrt {x}\right )^2-4=x-4.\] The formula looks like a straight line, but \(f\) must be defined first: \[\text {domain}=[0,\infty ),\] and as \(x\) runs from \(0\) upwards, \(x-4\) runs from \(-4\) upwards: \[\text {range}=[-4,\infty ).\]

Solution. First find the two inverses, by writing \(y\) for the output and making \(x\) the subject. \[y=2x+3\ \implies \ x=\frac {y-3}{2}\quad \implies \quad f^{-1}(x)=\frac {x-3}{2},\] \[y=3x-5\ \implies \ x=\frac {y+5}{3}\quad \implies \quad g^{-1}(x)=\frac {x+5}{3}.\]

(a)(i). Build the composite, then invert it: \[(f\circ g)(x)=f(3x-5)=2(3x-5)+3=6x-10+3=6x-7.\] \[y=6x-7\ \implies \ x=\frac {y+7}{6}\quad \implies \quad (f\circ g)^{-1}(x)=\frac {x+7}{6}.\] \[\therefore \quad (f\circ g)^{-1}(2)=\frac {2+7}{6}=\frac {9}{6}=\frac {3}{2}.\]

(ii). Work from the inside out. \(g^{-1}\) acts first: \[g^{-1}(4)=\frac {4+5}{3}=\frac {9}{3}=3,\] \[f^{-1}(3)=\frac {3-3}{2}=0.\] \[\therefore \quad \left (f^{-1}\circ g^{-1}\right )(4)=0.\]

(iii). Apply \(g^{-1}\) twice: \[g^{-1}(-10)=\frac {-10+5}{3}=\frac {-5}{3},\] \[g^{-1}\left (-\tfrac {5}{3}\right )=\frac {-\frac {5}{3}+5}{3}=\frac {\frac {10}{3}}{3} =\frac {10}{9}.\] \[\therefore \quad \left (g^{-1}\circ g^{-1}\right )(-10)=\frac {10}{9}.\]

(b). Build the composite: \[(g\circ f)(x)=g(x-4)=\frac {3}{(x-4)+1}=\frac {3}{x-3}.\] Now the equation \((g\circ f)^{-1}(x)=6\) says that the inverse sends \(x\) to \(6\). Reading that the other way round, the function itself sends \(6\) to \(x\): \[x=(g\circ f)(6)=\frac {3}{6-3}=\frac {3}{3}=1.\]

\[\therefore \quad x=1.\]

Solution. The test is to compute \(f(-x)\) and compare it with \(f(x)\): \[f(-x)=f(x)\ \implies \ \text {even},\qquad f(-x)=-f(x)\ \implies \ \text {odd},\] and if it is neither, the function is neither.

(a). Odd powers change sign, so \[f(-x)=4(-x)^5-7(-x)^3+(-x)=-4x^5+7x^3-x=-\left (4x^5-7x^3+x\right )=-f(x).\] Odd.

(b). \[f(-x)=(-x)^4-(-x)=x^4+x.\] This is not \(f(x)=x^4-x\), and it is not \(-f(x)=-x^4+x\) either. Neither.

(c). Both \(x^2\) terms are unaffected by the sign change: \[f(-x)=\frac {(-x)^2}{(-x)^2+1}=\frac {x^2}{x^2+1}=f(x).\] Even.

(d). The numerator changes sign and the denominator does not: \[f(-x)=\frac {-x}{(-x)^2+1}=\frac {-x}{x^2+1}=-f(x).\] Odd.

(e). \[f(-x)=\frac {1+(-x)}{1-(-x)}=\frac {1-x}{1+x}.\] Comparing with \(f(x)=\frac {1+x}{1-x}\): the numerator and denominator have swapped, which is neither leaving the function alone nor negating it. Neither.