The function h is defined by $$h(x) = \frac{\sqrt{x}}{x - 3}$$ where h has its maximum possible domain - AQA - A-Level Maths Mechanics - Question 10 - 2021 - Paper 2

Alice calculates:

• $h(1) = -0.5$
• $h(4) = 2$.

She observes a change of sign between these two points and concludes there must be a root of (h(x) = 0) in the interval ( (1, 4) ). However, the error lies in the fact that a change of sign does not guarantee a root if the function is not continuous throughout the interval.

Specifically, (h(x)) is not continuous at (x = 3), which lies between 1 and 4. Since (x = 3) is where the function has a discontinuity, we cannot definitively conclude there is a root based solely on sign change due to the discontinuity affecting the function’s behavior in that interval.

To determine whether h has an inverse function, we need to examine its turning points:

1. Differentiate h: $h(x) = \frac{\sqrt{x}}{x - 3}$ To differentiate, we use the quotient rule:

   $h'(x) = \frac{(x - 3)(\frac{1}{2\sqrt{x}}) - \sqrt{x}(1)}{(x - 3)^2}$

   Simplifying gives us: $h'(x) = \frac{\frac{x - 3}{2\sqrt{x}} - \sqrt{x}}{(x - 3)^2} = \frac{x - 3 - 2x}{2\sqrt{x}(x - 3)^2} = \frac{-x - 3}{2\sqrt{x}(x - 3)^2}$

   Setting (h'(x) = 0) yields: $-x - 3 = 0 \Rightarrow x = -3$ (not valid since domain is (x \geq 0)). Therefore, consider the interval ((0, \infty)) excluding 3.

2. Behavior of h'(x): Analyze when (h'(x) > 0) or (h'(x) < 0). Since the only critical point in the domain is (3) (a vertical asymptote), we find that:

   • As (x \to 0^+), (h(x) \to 0) (negative), and as (x \to 3^-), (h(x) \to -\infty) (negative).
   • As (x \to 3^+), (h(x) \to +\infty) (positive) then eventually approaches zero again as (x \to +\infty).

Given that there is a discontinuity and no turning points in the graph, h is not a one-to-one function throughout its domain. Therefore, h does not have an inverse function. A function must be one-to-one (i.e., no repeating y-values) to have an inverse.