9. (a) Show that f'(x) = \frac{(3 - x^2)^2}{x^2}, \quad x \neq 0 where A and B are constants to be found - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 1

To find ( f''(x) ), we first need to differentiate ( f'(x) ):

Starting from:

$$f'(x) = 9x^{-2} - 6 + x^2$$

We differentiate:

$$f''(x) = -18x^{-3} + 2x$$

We know ( f(-3) = 10 ). We can integrate ( f'(x) ) to find ( f(x) ):

Starting from:

$$f'(x) = 9x^{-2} - 6 + x^2$$

Integrating term by term:

$$f(x) = \int (9x^{-2} - 6 + x^2) dx = -9x^{-1} - 6x + \frac{x^3}{3} + c$$

Using the point (-3, 10):

$$10 = -9(-\frac{1}{3}) - 6(-3) + \frac{(-3)^3}{3} + c,$$

Calculating the left-hand side:

$$= 3 + 18 - 9 + c = 10 + c \quad \Rightarrow \quad c = -2$$

Thus, we have:

f(x) = -9x^{-1} - 6x + \frac{x^3}{3} - 2$$