The function h is defined by $$h(x) = \frac{\sqrt{x}}{x - 3}$$ where h has its maximum possible domain. 10 (a) Find the domain of h. Give your answer using set notation. 10 (b) Alice correctly calculates $$h(1) = -0.5 \text{ and } h(4) = 2$$ She then argues that since there is a change of sign there must be a value of x in the interval $1 < x < 4$ that gives $h(x) = 0$ Explain the error in Alice's argument. 10 (c) By considering any turning points of h, determine whether h has an inverse function. Fully justify your answer.

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

Question 10

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The function h is defined by $$h(x) = \frac{\sqrt{x}}{x - 3}$$ where h has its maximum possible domain. 10 (a) Find the domain of h. Give your answer using set n... show full transcript

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

Step 1

Find the domain of h.

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Answer

To find the domain of the function $h(x) = \frac{\sqrt{x}}{x - 3}$ , we need to consider the conditions under which the function is defined:

The expression inside the square root, $x$ , must be non-negative, which gives: $x \geq 0$
The denominator, $x - 3$ , must not be zero, which gives: $x \neq 3$

Taking both conditions into account, we can express the domain of h in set notation as: $\{ x \in \mathbb{R} : x \geq 0 \text{ and } x \neq 3 \}$

Step 2

Explain the error in Alice's argument.

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Answer

Alice's calculation of $h(1)$ and $h(4)$ is correct; however, her reasoning contains a flaw. The function $h(x)$ is not continuous at $x = 3$ , which makes her conclusion invalid. The change of sign between $h(1) = -0.5$ and $h(4) = 2$ does not guarantee the existence of a root in the interval $(1, 4)$ because the function is discontinuous at $x = 3$ . Hence, there is no guarantee that there is a value of $x$ such that $h(x) = 0$ in that interval.

Step 3

By considering any turning points of h, determine whether h has an inverse function.

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Answer

To determine whether the function h has an inverse, we first calculate the derivative: $h'(x) = \frac{d}{dx}\left(\frac{\sqrt{x}}{x - 3}\right).$ Using the quotient rule, we find: $h'(x) = \frac{(x - 3)\cdot\frac{1}{2\sqrt{x}} - \sqrt{x}\cdot 1}{(x - 3)^2} = \frac{\frac{x - 3}{2\sqrt{x}} - \sqrt{x}}{(x - 3)^2}.$

Setting this equal to zero for critical points, we need to solve: $\frac{x - 3 - 2x}{2\sqrt{x}} = 0,$ which simplifies to: $x - 3 - 2x = 0 \implies -x + 3 = 0 \implies x = 3.$

Since this point $x = 3$ is where the function is discontinuous, we identify it as a point where the function has no turning points. With no intervals where the function is either strictly increasing or strictly decreasing, there are no turning points.

As a result, the function does not pass the Horizontal Line Test across its domain, thus it is not one-to-one and does not have an inverse function.

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

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

Find the domain of h.

Explain the error in Alice's argument.

By considering any turning points of h, determine whether h has an inverse function.

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