Given that k is a negative constant and that the function f(x) is defined by f(x) = 2 - \frac{(x-5k)(x-k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0 (a) show that f(x) = \frac{x + k}{x - 2k} (b) Hence find f'(x), giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 3

To find the derivative of f(x), we use the quotient rule, which states:

$f'(x) = \frac{v u' - u v'}{v^2}$

where (u = x + k) and (v = x - 2k).

1. Differentiate u and v:

   • (u' = 1)
   • (v' = 1)

2. Substitute into the formula: $f'(x) = \frac{(x - 2k)(1) - (x + k)(1)}{(x - 2k)^2}$

3. Simplify:

   • Expand: (f'(x) = \frac{x - 2k - x - k}{(x - 2k)^2} = \frac{-3k}{(x - 2k)^2}$$

Thus, the simplified form of the derivative is:

$f'(x) = \frac{-3k}{(x - 2k)^2}$

To determine whether f(x) is increasing or decreasing, we analyze the sign of f'(x). Since k is a negative constant, we note:

1. Numerator: (-3k > 0), because multiplying a negative constant by -3 yields a positive result.
2. Denominator: ((x - 2k)^2 > 0) for all x not equal to 2k.

From this, we can conclude that:

• For all (x \geq 0), (f'(x) > 0). This indicates that f(x) is an increasing function.

In summary, we state that:

• f(x) is increasing because the derivative f'(x) is positive for all valid values of x.