Functions

Question 1

(a) Denominator zero at $x = -1$. Domain: $x \in \mathbb{R}, x \neq -1$. (1 exclusion, 1 statement)

(b) Solve $y = \frac{2x-3}{x+1}$ for $x$: $y(x+1) = 2x - 3 \Rightarrow yx + y = 2x - 3 \Rightarrow x(y-2) = -3 - y \Rightarrow x = \frac{-3-y}{y-2}$. This fails when $y = 2$. So range is $y \in \mathbb{R}, y \neq 2$ (the horizontal asymptote value). (algebra 2, statement 1)

(c) $f$ is a rational Möbius-type function; it is one-to-one on its domain (passes horizontal line test since inverting gives a unique $x$ for each $y$). From (b): $f^{-1}(x) = \frac{-3 - x}{x - 2} = \frac{-(x+3)}{x-2}, \quad x \neq 2.$ Domain of $f^{-1}$ = range of $f$ = $x \neq 2$. (one-to-one 1, inverse 2, domain 1)

(d) Setting $f(x) = f^{-1}(x)$; fixed points of $f$ satisfy $f(x) = x$: $\frac{2x-3}{x+1} = x \Rightarrow 2x - 3 = x^2 + x \Rightarrow x^2 - x + 3 = 0$, discriminant $1 - 12 < 0$ — no real fixed points. Solve $f(x)=f^{-1}(x)$ directly: $\frac{2x-3}{x+1} = \frac{-(x+3)}{x-2}$. Cross-multiply: $(2x-3)(x-2) = -(x+3)(x+1)$. LHS $= 2x^2 -7x +6$; RHS $= -(x^2+4x+3) = -x^2 -4x -3$. $2x^2 -7x +6 = -x^2 -4x -3 \Rightarrow 3x^2 -3x +9 = 0 \Rightarrow x^2 - x + 3 = 0$, no real solutions. No real solutions. (1)

Question 2

(a) $(g\circ h)(x) = g(h(x)) = \sqrt{(x^2-5)+4} = \sqrt{x^2 - 1}$. (2) Domain: need $x^2 - 1 \ge 0 \Rightarrow x \le -1$ or $x \ge 1$, i.e. $|x| \ge 1$. (2)

(b) $(h\circ g)(x) = h(g(x)) = (\sqrt{x+4})^2 - 5 = (x+4) - 5 = x - 1$. (2) Domain restricted by $g$: need $x + 4 \ge 0 \Rightarrow x \ge -4$. So domain $x \ge -4$. (1)

(c) They are not equal. Counterexample $x = 2$: $(g\circ h)(2) = \sqrt{4-1} = \sqrt3$, $(h\circ g)(2) = 2 - 1 = 1$. Since $\sqrt3 \neq 1$, claim false. (2)

(d) $(h\circ g)(x) = x - 1 = 0 \Rightarrow x = 1$. Check domain $x \ge -4$: valid. So $x = 1$. (2)

Question 3

(a) $p(x) = -f(x-1) + 3$: replace $x \to x+1$ (shift right 1), reflect in x-axis, shift up 3. Point $(a, b) \mapsto (a+1, -b+3)$.

• $(-2,5) \mapsto (-1, -2)$
• $(0,1) \mapsto (1, 2)$
• $(3,-2) \mapsto (4, 5)$ (1 each + 1 method = 4)

(b) Order: (1) horizontal shift right by 1 unit; (2) reflection in the x-axis; (3) vertical shift up by 3 units. (3, 1 each)

(c) $q(x) = f(2x)$ is a horizontal compression by factor $\tfrac12$. Point $(-2,5)$: we need $2x = -2 \Rightarrow x = -1$, so the point that had $x=-2$ moves to $x=-1$, giving $(-1, 5)$. Graph is squeezed horizontally toward the y-axis. (compression 1, point 1, description 1)

Question 4

(a)

• $F(-3) = (-3)^2 = 9$ (uses $x<0$ branch)
• $F(0) = 2 - 0 = 2$ (uses $0\le x\le2$ branch)
• $F(2) = 2 - 2 = 0$ (uses $0\le x\le2$ branch)
• $F(5) = 5 - 2 = 3$ (uses $x>2$ branch) (1 each)

(b) Each $x$ maps to exactly one output (branches cover disjoint intervals), so any vertical line meets the graph once → it is a function. (2) Continuity at $x=0$: left limit $\to x^2 \to 0$; value $F(0) = 2$. Since $0 \neq 2$, there is a jump — $F$ is discontinuous at $x = 0$. (1)

(c)

• For $x<0$: $x^2$ is decreasing (as $x$ increases toward 0, values fall from large to 0) → decreasing on $(-\infty, 0)$.
• For $0\le x\le 2$: $2-x$ has negative slope → decreasing on $[0,2]$.
• For $x>2$: $x-2$ has positive slope → increasing on $(2,\infty)$. Increasing: $(2,\infty)$. Decreasing: $(-\infty,0)$ and $[0,2]$. (3)

Question 5

(a) (i) $u(-x) = -x^3 + 4x = -(x^3 - 4x) = -u(x)$ → odd. (2) (ii) $v(-x) = \frac{x^2}{x^4+1} = v(x)$ → even. (2) (iii) $w(-x) = x^2 - x$; not equal to $w(x) = x^2+x$ nor to $-w(x)$ → neither. (2)

(b) Odd: $s(-x) = -s(x)$. So $s(-3) = -s(3) = -7$. (1.5) At $x=0$: $s(0) = -s(0) \Rightarrow 2s(0)=0 \Rightarrow s(0) = 0$ (0 is in domain of odd function by symmetry assumption). (1.5)

[
  {"claim":"f inverse of Q1 gives f(f_inv(x))=x", "code":"x=symbols('x'); f=lambda t:(2*t-3)/(t+1); finv=lambda t:(-(t+3))/(t-2); result = simplify(f(finv(x))-x)==0"},
  {"claim":"Q1(d) f(x)=f_inv(x) reduces to x^2-x+3=0 with no real roots", "code":"x=symbols('x'); expr=Eq((2*x-3)*(x-2), -(x+3)*(x+1)); sols=solve(expr,x); result = all(not s.is_real for s in sols)"},
  {"claim":"Q2 g(h(x))=sqrt(x^2-1)", "code":"x=symbols('x'); result = simplify(sqrt((x**2-5)+4) - sqrt(x**2-1))==0"},
  {"claim":"Q2 h(g(x))=x-1", "code":"x=symbols('x'); result = simplify((sqrt(x+4))**2 -5 - (x-1))==0"},
  {"claim":"Q5(i) u is odd", "code":"x=symbols('x'); u=x**3-4*x; result = simplify(u.subs(x,-x)+u)==0"},
  {"claim":"Q5(ii) v is even", "code":"x=symbols('x'); v=x**2/(x**4+1); result = simplify(v.subs(x,-x)-v)==0"}
]