The curve C has equation y = f(x) where f(x) = \frac{4x + 1}{x - 2}, \quad x > 2 (a) Show that f'(x) = \frac{-9}{(x - 2)^2} - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 5

We know that:

$f'(x) = \frac{-9}{(x - 2)^2}$

Setting this equal to -1:

$\frac{-9}{(x - 2)^2} = -1$

Multiplying both sides by ( (x - 2)^2 ):

$-9 = -(x - 2)^2$

Thus:

$(x - 2)^2 = 9$

Taking the square root of both sides gives:

$x - 2 = 3 \quad \text{or} \quad x - 2 = -3$

This simplifies to:

$x = 5 \quad \text{or} \quad x = -1$

As ( x > 2 ), we take ( x = 5 ).

Now, we find the coordinates by substituting ( x = 5 ) into the original function:

$f(5) = \frac{4(5) + 1}{5 - 2} = \frac{20 + 1}{3} = \frac{21}{3} = 7$

Thus, the coordinates of point P are ( (5, 7) ).