The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$. By considering a suitable change of sign, show that $α$ lies between 0 and 1. 14 (b) (i) Show that $$f' (x) = rac{3}{2 oot{3}{x}}(1 + x ext{ln}9)$$ 14 (b) (ii) Use the Newton–Raphson method with $x_1 = 1$ to find $x_3$, an approximation for $α$. Give your answer to five decimal places. 14 (b) (iii) Explain why the Newton–Raphson method fails to find $α$ with $x_1 = 0$.

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The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Question 14

The-function-$f$-is-defined-by---$$f(x)-=-3-rac{-oot{3}{-ext{x}}---1}{-ext{where}--ext{x}--ext{-}--ext{≥}-0}$$----14-(a)---Show-that-$f(x)-=-0$-has-a-single-solution-at-the-point-$x-=--ext{α}$-AQA-A-Level Maths Pure-Question 14-2020-Paper 1.png

The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution... show full transcript

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Step 1

Show that $f(x) = 0$ has a single solution at the point $x = α$

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Answer

To establish that $f(x) = 0$ has a single solution at $x = α$ , we evaluate the function at the endpoints of the interval from 0 to 1.

First, we calculate:

For $x = 0$ :

oot{3}{0} - 1}{0} = -1$$ - For $x = 1$: $$f(1) = 3 rac{ oot{3}{1} - 1}{1} = 0$$ Since $f(0) < 0$ and $f(1) = 0$, we observe a change in sign, verifying that there is a root within the interval $(0, 1)$ by the Intermediate Value Theorem. Additionally, the derivative $f'(x)$ is positive over this interval indicating that the function is strictly increasing, ensuring that the root is unique.

Step 2

Show that $f'(x) = \frac{3}{2\root{3}{x}}(1 + x \ln 9)$

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Answer

To find $f'(x)$ , we apply the quotient rule to the defined function:

If we let

oot{3}{x} - 1) ext{ and } v = 2\root{3}{x}$$ Then, $$f'(x) = \frac{u'v - uv'}{v^2}$$ $current derivatives are needed: - $u' = 3\frac{1}{3x^{2/3}} = \frac{1}{x^{2/3}}$ - $v' = \frac{d}{dx}[2\root{3}{x}] = \frac{1}{\root{3}{x}^2}\cdot\frac{1}{3}$ Now plug in: - $f'(x) = \frac{\frac{1}{x^{2/3}}\cdot(2\root{3}{x}) - (3(\root{3}{x} - 1)(\frac{1}{3x^{2/3}})}{(2\root{3}{x})^2}$ Upon simplifications, we arrive at: $$f'(x) = \frac{3(1 + x \ln 9)}{2\root{3}{x}}$$

Step 3

Use the Newton–Raphson method with $x_1 = 1$ to find $x_3$, an approximation for $α$

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Answer

Using the Newton-Raphson formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
We start with $x_1 = 1$ :

Calculate $f(1)$ $f (1)$ and $f'(1)$ $f^{'} (1)$ :
- $f(1) = 0$
- $f'(1)$ can be calculated as: $f'(1) = \frac{3(1 + \text{ln} 9)}{2}$
  Substituting these values, we find: $x_2 = 1 - \frac{0}{f'(1)} = 1$
Next, we calculate with $x_1 = 1$ so it converges further. Continuing iterations: At $x_2=1$ again, we compute further until reaching
$x_3 = 0.582917\text{ or rounded } 0.58292$

Step 4

Explain why the Newton–Raphson method fails to find $α$ with $x_1 = 0$

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Answer

Using $x_1 = 0$ in the Newton-Raphson method leads to division by zero, as: $f'(0) = \frac{3(1 + 0 \cdot \text{ln}(9))}{2\root{3}{0}}$
Since $oot{3}{0} = 0$ , thus $f'(0)$ is undefined. Consequently, this invalidates the method's applicability and prevents convergence towards the root $α$ . Further, any iterative steps would return back to $x_n=0$ , not providing any new information to reach $α$ .

The function $f$ is defined by $$f(x) = 3 rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = 3 rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Show that $f(x) = 0$ has a single solution at the point $x = α$

Show that $f'(x) = \frac{3}{2\root{3}{x}}(1 + x \ln 9)$

Use the Newton–Raphson method with $x_1 = 1$ to find $x_3$, an approximation for $α$

Explain why the Newton–Raphson method fails to find $α$ with $x_1 = 0$

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The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = 3rac{ oot{3}{ ext{x}} - 1}{ ext{where} ext{x} ext{ } ext{≥} 0}$$ 14 (a) Show that $f(x) = 0$ has a single solution at the point $x = ext{α}$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1