Functions — practice questions

(a) To complete the square: $f(x) = (x^2 - 6x) + 5$. We halve the coefficient of $x$, which is -3. So, $f(x) = (x-3)^2 - (-3)^2 + 5 = (x-3)^2 - 9 + 5$. Therefore, $f(x) = (x-3)^2 - 4$. [1 mark]

(b) The vertex of the parabola $y=f(x)$ is at $(3, -4)$. Since the coefficient of $x^2$ is positive, the parabola opens upwards. The minimum value of $f(x)$ is -4. The range is $f(x) \ge -4$. [1 mark]

(c) A function has an inverse only if it is one-to-one. The function $f(x)$ is a many-to-one function. To make it one-to-one, we must restrict its domain to one side of the line of symmetry, $x=3$. The smallest value for $k$ that achieves this is $k=3$. [1 mark]

(d) Let $y = g(x)$, so $y = (x-3)^2 - 4$ for $x \ge 3$. To find the inverse, we swap $x$ and $y$: $x = (y-3)^2 - 4$. Now, we make $y$ the subject: $x+4 = (y-3)^2$ $\sqrt{x+4} = y-3$ $y = 3 + \sqrt{x+4}$. We take the positive square root because the domain of $g(x)$ was $x \ge 3$, which means the range of $g^{-1}(x)$ must be $y \ge 3$. So, $g^{-1}(x) = 3 + \sqrt{x+4}$. [2 marks] The domain of $g^{-1}(x)$ is the range of $g(x)$. From part (b), the minimum value is -4. So, the range of $g(x)$ is $y \ge -4$. Therefore, the domain of $g^{-1}(x)$ is $x \ge -4$. [1 mark]

(a) $fg(x) = f(g(x))$. We substitute the expression for $g(x)$ into $f(x)$. $fg(x) = 2(g(x)) + 1 = 2\left(\frac{12}{x+2}\right) + 1$ [1 mark] $fg(x) = \frac{24}{x+2} + 1$. To simplify, we find a common denominator: $fg(x) = \frac{24}{x+2} + \frac{x+2}{x+2} = \frac{24 + x + 2}{x+2} = \frac{x+26}{x+2}$. [1 mark]

(b) To find the range of $y = \frac{x+26}{x+2}$, we can consider the horizontal asymptote. As $x \to \infty$, $y \to \frac{x}{x} = 1$. Alternatively, let's make $x$ the subject: $y(x+2) = x+26$ $yx + 2y = x+26$ $yx - x = 26 - 2y$ $x(y-1) = 26 - 2y$ $x = \frac{26-2y}{y-1}$. The expression for $x$ is undefined when the denominator is zero, i.e., when $y-1=0$ or $y=1$. Thus, $y$ can be any real number except 1. The range is $fg(x) \in \mathbb{R}, fg(x) \ne 1$. [2 marks]

(c) We set our expression for $fg(x)$ equal to 5: $\frac{x+26}{x+2} = 5$ [1 mark] $x+26 = 5(x+2)$ $x+26 = 5x+10$ $16 = 4x$ $x=4$. [1 mark] We check that this solution is valid. The domain of $fg(x)$ is $x \ne -2$. Since $4 \ne -2$, the solution is valid.