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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
The function h is defined by $$ h(x) = \frac{\sqrt{x}}{x - 3} $$ where h has its maximum possible domain. (a) Find the domain of h. Give your answer using set no... 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.
Answer
To find the domain of the function , we must ensure both the numerator and denominator are defined.
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Numerator Requirement: The expression ( \sqrt{x} ) requires that ( x \geq 0 ). This leads to the inequality.
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Denominator Requirement: The denominator ( x - 3 ) cannot be zero, thus ( x \neq 3 ).
Combining these conditions, we find the domain in set notation to be: .
Thus, the domain of h is: .
Step 2
Explain the error in Alice's argument.
Answer
Alice finds that ( h(1) = -0.5 ) and ( h(4) = 2 ), noting the change of sign between these two values. However, she assumes that a root must exist in this interval due to the Intermediate Value Theorem.
The key oversight is that ( h(x) ) is not continuous across all points in the interval ( (1, 4) ) since there is a discontinuity at ( x = 3 ). Therefore, the presence of a change in sign does not guarantee a root exists within this interval.
Step 3
By considering any turning points of h, determine whether h has an inverse function.
Answer
To determine if h has an inverse function, we need to consider the turning points. We first differentiate h:
Setting ( h'(x) = 0 ), we solve:
This simplifies to:
Since ( x = -3 ) is not within the domain of h, we note that there are no turning points.
As h does not change direction and is monotonic on its domain, it is one-to-one and hence has an inverse function.