Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = 8 \sin \left( \frac{1}{2} x \right) - 3x + 9 \quad (x > 0)$ and $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 7 - 2022 - Paper 2

To explain why $a$ lies in the interval $[4, 5]$, we can use the information provided about the function values:

1. Given that $f(4) = 4.274 > 0$ and $f(5) = -1.212 < 0$, we see:

   • At $x = 4$, the function is positive.
   • At $x = 5$, the function is negative.

2. By the Intermediate Value Theorem, since $f(x)$ is continuous, and there is a sign change between $x = 4$ and $x = 5$, there must be some value $a$ in the interval $[4, 5]$ such that $f(a) = 0$.

1. Using the Newton-Raphson formula:

   $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

2. We start with $x_0 = 5$:

   • First, calculate $f(5)$ and $f'(5)$:

   • Given previously:

   $f(5) = -1.212$ and $f'(5) = 4\cos\left(\frac{1}{2} \times 5\right) - 3 \approx 4\cos(2.5) - 3 \\$

   Approximating:

   $f'(5) \approx 0.996$ (calculated using a calculator).

3. Now apply the Newton-Raphson step:

   $x_1 = 5 - \frac{-1.212}{0.996} \approx 5 + 1.216 \approx 4.784$

4. Rounding to 3 significant figures gives:

   $x \approx 4.78$.