The function f is defined by $$f : x \mapsto \frac{2(x-1)}{x^2 - 2x - 3} + \frac{1}{x - 3}, \quad x > 3.$$ (a) Show that $$f(x) = \frac{1}{x + 1}, \quad x > 3.$$ (b) Find the range of f - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 5

The range of ( f ) can be determined by analyzing the expression for ( f(x) ):

1. We know from part (a) that: $f(x) = \frac{1}{x + 1}, \quad x > 3$

2. As ( x ) approaches 3 from the right, ( f(3) = \frac{1}{4} ).

3. As ( x ) increases towards infinity, ( f(x) ) approaches 0: $\lim_{x \to \infty} f(x) = 0$

4. Therefore, the range of ( f ) is: $\left(0, \frac{1}{4}\right)$

To find the inverse, let ( y = f(x) ): [ y = \frac{1}{x + 1} ]

1. Swap ( x ) and ( y ): [ x = \frac{1}{y + 1} ]

2. Rearranging gives: [ xy + x = 1 \quad \Rightarrow \quad xy = 1 - x \quad \Rightarrow \quad y = \frac{1 - x}{x} \] Thus, the inverse function is: f^{-1}(x) = \frac{1 - x}{x}$$

3. The domain of ( f^{-1}(x) ) must be aligned with the range of ( f ), therefore: [ Dom(f^{-1}) = \left(0, \frac{1}{4}\right) ]

To solve this equation:

1. Using the definition of ( g ): $g(x) = 2x - 3$

2. Substitute into ( f ): $fg(x) = f(2x - 3)$

3. We know: $f(x) = \frac{1}{x + 1}$ so: $f(2x - 3) = \frac{1}{(2x - 3) + 1} = \frac{1}{2x - 2}$

4. Set this equal to ( \frac{1}{8} ): $\frac{1}{2x - 2} = \frac{1}{8}$

5. Cross-multiply and solve: $8 = 2x - 2 \\ => 2x = 10 \\ => x = 5$

6. Check both conditions:

   • Check ( g(5) ): $g(5) = 2(5) - 3 = 7$
   • Verify with ( f(g(5)) ): $f(7) = \frac{1}{7 + 1} = \frac{1}{8}$ Hence, the solution is valid.