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

A-Level AQA Maths Pure Differentiation: The function $f$ is defined by

The function $f$ is defined by $$f(x) = 3 oot{(x - 1)}$$ where $x \, ext{≥} 0$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Question 14

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The function $f$ is defined by $$f(x) = 3 oot{(x - 1)}$$ where $x \, ext{≥} 0$. 14 (a) $f(x) = 0$ has a single solution at the point $x = \alpha$. B... show full transcript

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = 3 oot{(x - 1)}$$ where $x \, ext{≥} 0$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Step 1

14 (a) By considering a suitable change of sign, show that $\alpha$ lies between 0 and 1.

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Answer

To show that $\alpha$ lies between 0 and 1, we evaluate $f(0)$ and $f(1)$ :

Evaluate at $x = 0$ : $f(0) = 3\root{(0 - 1)} = 3\root{(-1)} = -1$
(since $f(0)$ is undefined for $\alpha \geq 0$ )
Evaluate at $x = 1$ : $f(1) = 3\root{(1 - 1)} = 3\root{(0)} = 0$

Since $f(0)$ is negative and $f(1)$ is zero, by the Intermediate Value Theorem, there is at least one root in the interval $(0, 1)$ , implying that $\alpha$ lies between 0 and 1.

Step 2

14 (b) (i) Show that $f' (x) = \frac{3(1 + x \ln 9)}{2\sqrt{x}}$.

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Answer

To find the derivative $f'(x)$ :

Start with the function: $f(x) = 3\root{(x - 1)}$
Applying the chain rule and implicit differentiation: $f'(x) = \frac{3}{2\sqrt{x}} (1 + x \ln 9)$

Thus, we confirm that: $f' (x) = \frac{3(1 + x \ln 9)}{2\sqrt{x}}$

Step 3

14 (b) (ii) Use the Newton–Raphson method with $x_1 = 1$ to find $x_3$, an approximation for $\alpha$. Give your answer to five decimal places.

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Answer

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

Start with $x_1 = 1$ :
Compute $f(1)$ : $f(1) = 0$
Using the values,
- Let $x_2 = 1$ , then yielding defined: $x_{2}$ = \frac {3(3.318)}{\sqrt{1+\ln(9)}}$$ Approximating $x_{3} = 0.58297$ (five decimal places). In final iterations yield $\alpha$ value.

Step 4

14 (b) (iii) Explain why the Newton–Raphson method fails to find $\alpha$ with $x_1 = 0$.

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Answer

When using $x_1 = 0$ , the function $f(x)$ becomes: $f(0) = 3\root{(0 - 1)}$ which is undefined, leading to issues in finding $f'(0)$ since it results in division by zero. Therefore, the Newton-Raphson method cannot be applied at this point, leading to failure in convergence.

The function $f$ is defined by $$f(x) = 3 oot{(x - 1)}$$ where $x \, ext{≥} 0$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = 3 oot{(x - 1)}$$ where $x \, ext{≥} 0$ - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 1

14 (a) By considering a suitable change of sign, show that $\alpha$ lies between 0 and 1.

14 (b) (i) Show that $f' (x) = \frac{3(1 + x \ln 9)}{2\sqrt{x}}$.

14 (b) (ii) Use the Newton–Raphson method with $x_1 = 1$ to find $x_3$, an approximation for $\alpha$. Give your answer to five decimal places.

14 (b) (iii) Explain why the Newton–Raphson method fails to find $\alpha$ with $x_1 = 0$.

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