A function $f$ is defined by
$$f(x) = \frac{-x}{\sqrt{2x - 2}}$$ - AQA - A-Level Maths Mechanics - Question 6 - 2018 - Paper 3
Question 6
A function $f$ is defined by
$$f(x) = \frac{-x}{\sqrt{2x - 2}}$$.
6 (a) State the maximum possible domain of $f$.
6 (b) Use the quotient rule to show that $f'(x) ... show full transcript
Worked Solution & Example Answer:A function $f$ is defined by
$$f(x) = \frac{-x}{\sqrt{2x - 2}}$$ - AQA - A-Level Maths Mechanics - Question 6 - 2018 - Paper 3
Step 1
State the maximum possible domain of f.
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Answer
To determine the maximum possible domain of the function f(x)=2x−2−x, we need to ensure that the expression under the square root is non-negative.
Set the expression inside the square root greater than zero:
2x−2>0
Solving this inequality:
2x>2⟹x>1
Since the square root is in the denominator, we must also ensure that it is not zero:
2x−2=0⟹x=1
Therefore, combining these conditions, the maximum possible domain of f is:
x>1, or in interval notation: (1,∞).
Step 2
Use the quotient rule to show that f'(x) = \frac{-x - 2}{(2x - 2)^{\frac{3}{2}}}
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Answer
To differentiate the function f(x)=2x−2−x, we will use the quotient rule, which states:
(vu)′=v2u′v−uv′
where u=−x and v=2x−2.
First, find the derivatives u′ and v′:
u′=−1
To differentiate v=(2x−2)21, use the chain rule:
v′=21(2x−2)−21⋅2=2x−21
Substitute into the quotient rule formula:
f′(x)=(2x−2)2(−1)⋅2x−2−(−x)⋅2x−21
Simplifying the numerator:
f′(x)=2x−2−2x−2+2x−2x=2x−2−2x−2+2x−2x=2x−2−2x−2(2x−2)+x=(2x−2)23−x−2
