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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. 10 (a) Find the domain of h. Give your answer using se... 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 ( h(x) = \frac{\sqrt{x}}{x - 3} ), we need to consider two main conditions:
- Square Root Condition: The expression under the square root must be non-negative. Thus, ( x \geq 0 ).
- Denominator Condition: The denominator must not equal zero. Therefore, ( x - 3 \neq 0 ) implies ( x \neq 3 ).
Combining these conditions, the domain can be expressed in set notation as:
Step 2
Explain the error in Alice's argument.
Answer
Alice's argument contains a misunderstanding regarding continuity and the Intermediate Value Theorem. She correctly calculates ( h(1) = -0.5 ) and ( h(4) = 2 ), noting a change of sign, which seems to imply that there exists some ( x ) in (1, 4) where ( h(x) = 0 ". However, the function has a discontinuity at ( x = 3 ). Therefore, since the values ( h(1) ) and ( h(4) ) are actually on separate intervals (as ( x = 3 ) is not included), the Intermediate Value Theorem does not apply here. A change of sign does not guarantee the existence of a root.
Step 3
By considering any turning points of h, determine whether h has an inverse function.
Answer
To determine if the function ( h ) has an inverse, we must analyze its turning points by calculating the derivative. Differentiating ( h(x) ):
We find turning points by setting ( h'(x) = 0 ). This derivative can be complex to simplify; however, it reveals that turning points occur where the derivative changes sign. If the derivative does not have turning points (i.e., the function is monotonous), ( h ) would be one-to-one and thus have an inverse.
After testing intervals, we conclude that ( h ) is monotonically decreasing when ( x > 3 ) and does not change around the discontinuity, establishing that ( h ) has no turning points in its domain. Therefore, it is one-to-one, and:
Conclusion: ( h ) has an inverse function.