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The curve C has equation y = f(x) where f(x) = (4x + 1) / (x - 2) and x>2. Given that P is a point on C such that f'(x) = -1.

Firstly, in order to solve this problem we would need to differentiate f(x) to get f'(x).

To differentiate this we would use the quotient rule. The quotient rule is that:

dy/dx = (V.dU/dx - U.dX/dx) / V^2

where U = the numerator = 4x + 1

and V = the denominator = x - 2

This would give the result of:

dy/dx = ((x - 2)4 - (4x + 1)1) / (x - 2)^2

This would then cancel down to give

dy/dx = -9 / (x - 2)^2

Knowing that dy/dx is equivalent to f'(x), we can eqwuate our expression for dy/dx to the value given in the question for f-(x), which is -1.

-1 = -9 / (x - 2)^2

At this point we can solve for x. Firstly by expanding the bracket.

-1 = -9 / (x^2 - 4x + 4)

From this we can bring the denominator to the top and group all the terms on one side.

-1 (x^2 - 4x + 4) = -9

-x^2 + 4x -4 = -9

x^2 - 4x -5 =0

Now we can solve to find the x coordinates:

(x + 1) (x -5) = 0

giving that x = -1 and x = 5

We can substitute these x values into our equation for f(x) to get the corresponding y values.

There when x = -1

y = f(x) = (4(-1) + 1) / ((-1) - 2)

Giving that y = 1 when x = -1, thus the coordinates are (-1,1)

And when x = 5:

y = f(x) = (4(5) + 1) / ((5) - 2)

so y = 7 when x = 5, thus the coordinates are (5,7)

However, in the question we were given the limit x>2, meaning that the answer cannot be (-1,1) and thus the final answer is (5,7).

Chantelle C. GCSE Maths tutor, 11 Plus Maths tutor, A Level Maths tut...

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