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A function $f$ is defined by $$f(x) = \frac{-x}{\sqrt{2x - 2}}$$ 6 (a) State the maximum possible domain of $f$ - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 3

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A-function-$f$-is-defined-by---$$f(x)-=-\frac{-x}{\sqrt{2x---2}}$$--6-(a)-State-the-maximum-possible-domain-of-$f$-AQA-A-Level Maths Pure-Question 6-2018-Paper 3.png

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 $$f(x) = \frac{-x}{\sqrt{2x - 2}}$$ 6 (a) State the maximum possible domain of $f$ - AQA - A-Level Maths Pure - 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)=−x2x−2f(x) = \frac{-x}{\sqrt{2x-2}}, we need to ensure that the expression inside the square root is non-negative and that the denominator does not equal zero.

  1. The term under the square root is 2x−2≥02x - 2 \geq 0.
    • Solving this inequality:
    x \geq 1$$
  2. the square root must not equal zero:
    • 2x−2≠02x - 2 \neq 0 gives us the same condition, x≠1x \neq 1.

Thus, the maximum possible domain is x>1x > 1, or in interval notation, the domain is (1,∞)(1, \infty).

Step 2

Use the quotient rule to show that $f'(x) = \frac{-x - 2}{(2x - 2)^{3/2}}$.

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Answer

To find the derivative of the function f(x)=−x2x−2f(x) = \frac{-x}{\sqrt{2x-2}}, we will use the quotient rule, which states that if we have a function uv\frac{u}{v}, the derivative is given by:

f′(x)=u′v−uv′v2f'(x) = \frac{u'v - uv'}{v^2}

In our case, let:

  • u=−xu = -x
  • v=2x−2v = \sqrt{2x - 2}

Calculating derivatives:

  • u′=−1u' = -1
  • v=(2x−2)1/2  ⟹  v′=12(2x−2)−1/2⋅2=12x−2v = (2x - 2)^{1/2} \implies v' = \frac{1}{2}(2x - 2)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x - 2}}

Now applying the quotient rule:

f′(x)=(−1)2x−2−(−x)(12x−2)(2x−2)2f'(x) = \frac{(-1)\sqrt{2x - 2} - (-x)(\frac{1}{\sqrt{2x - 2}})}{(\sqrt{2x - 2})^2}

This simplifies to: f′(x)=−2x−2+x2x−22x−2f'(x) = \frac{-\sqrt{2x - 2} + \frac{x}{\sqrt{2x - 2}}}{2x - 2}

Combining the terms in the numerator: =−2x−2+x2x−2(2x−2)=−x−2(2x−2)3/2= \frac{-\sqrt{2x - 2} + \frac{x}{\sqrt{2x - 2}}}{(2x - 2)} = \frac{-x - 2}{(2x - 2)^{3/2}}

Thus, we have shown that: f′(x)=−x−2(2x−2)3/2f'(x) = \frac{-x - 2}{(2x - 2)^{3/2}}.

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