Unit 4: Functions and Graphs – Exercise 4.1

(i) \(f(x) + g(x)\) \[ \begin{aligned} f(x) + g(x) &= (2x + 5) + (3x - 2) \\ &= 2x + 5 + 3x - 2 \\ &= 5x + 3 \end{aligned} \]

(ii) \(f(x) - g(x)\) \[ \begin{aligned} f(x) - g(x) &= (2x + 5) - (3x - 2) \\ &= 2x + 5 - 3x + 2 \\ &= -x + 7 \end{aligned} \]

(i) \(f(x) \cdot g(x)\) \[ \begin{aligned} f(x) \cdot g(x) &= (x + 2)(2x + 4) \\ &= x(2x + 4) + 2(2x + 4) \\ &= 2x^2 + 4x + 4x + 8 \\ &= 2x^2 + 8x + 8 \\ &= 2(x^2 + 4x + 4) \end{aligned} \]

(ii) \(g(x) \cdot f(x)\) \[ \begin{aligned} g(x) \cdot f(x) &= (2x + 4)(x + 2) \\ &= 2x(x + 2) + 4(x + 2) \\ &= 2x^2 + 4x + 4x + 8 \\ &= 2x^2 + 8x + 8 \\ &= 2(x^2 + 4x + 4) \end{aligned} \]

(iii) \(\frac{f(x)}{g(x)}\) \[ \begin{aligned} \frac{f(x)}{g(x)} &= \frac{x + 2}{2x + 4} \\ &= \frac{x + 2}{2(x + 2)} = \frac{1}{2} \end{aligned} \]

(iv) \(\frac{g(x)}{f(x)}\) \[ \begin{aligned} \frac{g(x)}{f(x)} &= \frac{2x + 4}{x + 2} \\ &= \frac{2(x + 2)}{x + 2} = 2 \end{aligned} \]

(i) \(f(x) = 2x + 3,\; g(x) = x^3\)
\((f \circ g)(x) = f(g(x)) = f(x^3) = 2x^3 + 3\)
\((g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^3\)
\((f \circ f)(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 9\)
\((g \circ g)(x) = g(g(x)) = g(x^3) = (x^3)^3 = x^9\)

(ii) \(f(x) = \frac{2}{x},\; x \neq 0;\quad g(x) = 2x^2 - 1\)
\((f \circ g)(x) = f(g(x)) = f(2x^2 - 1) = \frac{2}{2x^2 - 1}\)
\((g \circ f)(x) = g(f(x)) = g\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 - 1 = 2\left(\frac{4}{x^2}\right) - 1 = \frac{8}{x^2} - 1 = \frac{8 - x^2}{x^2}\)
\((f \circ f)(x) = f(f(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = 2 \times \frac{x}{2} = x\)
\((g \circ g)(x) = g(g(x)) = g(2x^2 - 1) = 2(2x^2 - 1)^2 - 1 = 2[4x^4 + 1 - 4x^2] - 1 = 8x^4 + 2 - 8x^2 - 1 = 8x^4 - 8x^2 + 1\)

(iii) \(f(x) = 2x - 1,\; x \neq 0;\quad g(x) = \frac{x+1}{2}\)
\((f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{2}\right) = 2\left(\frac{x+1}{2}\right) - 1 = x + 1 - 1 = x\)
\((g \circ f)(x) = g(f(x)) = g(2x - 1) = \frac{(2x - 1) + 1}{2} = \frac{2x}{2} = x\)
\((f \circ f)(x) = f(f(x)) = f(2x - 1) = 2(2x - 1) - 1 = 4x - 2 - 1 = 4x - 3\)
\((g \circ g)(x) = g(g(x)) = g\left(\frac{x+1}{2}\right) = \frac{\frac{x+1}{2} + 1}{2} = \frac{\frac{x+1+2}{2}}{2} = \frac{x+3}{2} \times \frac{1}{2} = \frac{x+3}{4}\)

\[ \begin{aligned} (f \circ g)(x) &= f(g(x)) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2 \\ (g \circ f)(x) &= g(f(x)) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k \\ \text{Given: } & 18x - 3k + 2 = 18x + 12 - k \\ & -3k + 2 = 12 - k \\ & -3k + k = 12 - 2 \\ & -2k = 10 \\ & k = -5 \end{aligned} \]

(i) \(g(f(4))\) \[ \begin{aligned} f(4) &= 3(4) + 2 = 12 + 2 = 14 \\ g(f(4)) &= g(14) = 2(14) + 3 = 28 + 3 = 31 \end{aligned} \]

(ii) \(f(f(3))\) \[ \begin{aligned} f(3) &= 3(3) + 2 = 9 + 2 = 11 \\ f(f(3)) &= f(11) = 3(11) + 2 = 33 + 2 = 35 \end{aligned} \]

(iii) \(f(g(-2))\) \[ \begin{aligned} g(-2) &= 2(-2) + 3 = -4 + 3 = -1 \\ f(g(-2)) &= f(-1) = 3(-1) + 2 = -3 + 2 = -1 \end{aligned} \]

(i) \(f(x) = 2x - 3\) \[ \begin{aligned} y &= 2x - 3 \implies y + 3 = 2x \implies x = \frac{y+3}{2} \\ f^{-1}(y) &= \frac{y+3}{2} \implies f^{-1}(x) = \frac{x+3}{2} \end{aligned} \]

(ii) \(f(x) = 4x^3 - 1\) \[ \begin{aligned} y &= 4x^3 - 1 \implies y+1 = 4x^3 \implies x^3 = \frac{y+1}{4} \\ x &= \sqrt[3]{\frac{y+1}{4}} \implies f^{-1}(x) = \sqrt[3]{\frac{x+1}{4}} \end{aligned} \]

(iii) \(f(x) = \sqrt{x-1},\; x \ge 1\) \[ \begin{aligned} y &= \sqrt{x-1} \implies y^2 = x-1 \implies x = y^2 + 1 \\ f^{-1}(x) &= x^2 + 1 \end{aligned} \]

(iv) \(f(x) = \frac{x+1}{3x-2},\; x \neq \frac{2}{3}\) \[ \begin{aligned} y &= \frac{x+1}{3x-2} \implies y(3x-2) = x+1 \\ 3xy - 2y &= x+1 \implies 3xy - x = 2y + 1 \\ x(3y - 1) &= 2y + 1 \implies x = \frac{2y+1}{3y-1} \\ f^{-1}(x) &= \frac{2x+1}{3x-1} \end{aligned} \]

\(f^{-1}(x)\): \[ \begin{aligned} y &= 4x + 2 \implies y-2 = 4x \implies x = \frac{y-2}{4} \\ f^{-1}(x) &= \frac{x-2}{4} \end{aligned} \] \(g^{-1}(x)\): \[ \begin{aligned} y &= 6x - 18 \implies y+18 = 6x \implies x = \frac{y+18}{6} \\ g^{-1}(x) &= \frac{x+18}{6} \end{aligned} \] Solve \(f^{-1}(x) = g^{-1}(x)\): \[ \begin{aligned} \frac{x-2}{4} &= \frac{x+18}{6} \implies 6(x-2) = 4(x+18) \\ 6x - 12 &= 4x + 72 \implies 2x = 84 \implies x = 42 \end{aligned} \]

(i) \(f(x) = x - 6\) \[ \begin{aligned} y = x-6 &\implies x = y+6 \implies f^{-1}(x) = x+6 \\ f(f^{-1}(x)) &= f(x+6) = (x+6)-6 = x \\ f^{-1}(f(x)) &= f^{-1}(x-6) = (x-6)+6 = x \end{aligned} \]

(ii) \(f(x) = 7x - 4\) \[ \begin{aligned} y = 7x-4 &\implies y+4 = 7x \implies x = \frac{y+4}{7} \implies f^{-1}(x) = \frac{x+4}{7} \\ f(f^{-1}(x)) &= f\left(\frac{x+4}{7}\right) = 7\cdot\frac{x+4}{7} - 4 = x+4-4 = x \\ f^{-1}(f(x)) &= f^{-1}(7x-4) = \frac{(7x-4)+4}{7} = \frac{7x}{7} = x \end{aligned} \]

(iii) \(f(x) = \frac{x-4}{x+2},\; x \neq -2\) \[ \begin{aligned} y &= \frac{x-4}{x+2} \implies y(x+2) = x-4 \implies xy + 2y = x-4 \\ xy - x &= -4 - 2y \implies x(y-1) = -4-2y \implies x = \frac{-4-2y}{y-1} \\ f^{-1}(x) &= \frac{-4-2x}{x-1} \\ f(f^{-1}(x)) &= f\left(\frac{-4-2x}{x-1}\right) = \frac{\frac{-4-2x}{x-1} - 4}{\frac{-4-2x}{x-1} + 2} = \frac{-4-2x -4(x-1)}{-4-2x +2(x-1)} = \frac{-6x}{-6} = x \\ f^{-1}(f(x)) &= f^{-1}\left(\frac{x-4}{x+2}\right) = \frac{-4 - 2\left(\frac{x-4}{x+2}\right)}{\left(\frac{x-4}{x+2}\right) - 1} = \frac{-4(x+2)-2(x-4)}{x-4 - (x+2)} = \frac{-6x}{-6} = x \end{aligned} \]

(i) \(f(x) = 12x - 3\)
Domain of \(f\): \((-\infty, \infty)\)    Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\)    Range of \(f^{-1}\): \((-\infty, \infty)\)

(ii) \(f(x) = \frac{1}{2x+8}\)
Domain of \(f\): \((-\infty, \infty)\)    Range of \(f\): \((-\infty, \infty)\)
Domain of \(f^{-1}\): \((-\infty, \infty)\)    Range of \(f^{-1}\): \((-\infty, \infty)\)

(iii) \(f(x) = \frac{x}{1+x},\; x \neq -1\)
Domain of \(f\): \((-\infty, -1) \cup (-1, \infty)\)
Range of \(f\): \((-\infty, 1) \cup (1, \infty)\) (since solving \(y = \frac{x}{1+x} \implies x = \frac{y}{1-y}, y\neq1\))
Domain of \(f^{-1}\): \((-\infty, 1) \cup (1, \infty)\)    Range of \(f^{-1}\): \((-\infty, -1) \cup (-1, \infty)\)

(iv) \(f(x) = \sqrt{x-2},\; x \ge 2\)
Domain of \(f\): \([2, \infty)\)
Range of \(f\): \([0, \infty)\) (since square root yields non-negative values)
Domain of \(f^{-1}\): \([0, \infty)\)    Range of \(f^{-1}\): \([2, \infty)\)

\[ \begin{aligned} f(a) &= g(a) \\ a^2 + 9 &= a + 21 \\ a^2 - a - 12 &= 0 \\ a^2 - 4a + 3a - 12 &= 0 \\ a(a-4) + 3(a-4) &= 0 \\ (a-4)(a+3) &= 0 \\ a = 4 \quad \text{or} \quad a = -3 \end{aligned} \]