A-Level Edexcel Maths Pure Solving Equations: The curve C has equation y = f(x

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} (3) Given that P is a point on C such that f'(x) = -1, (b) find the coordinates of P - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 5

Question 2

$The-curve-C-has-equation-y-=-f(x)-where--f(x)-=-\frac{4x-+-1}{x---2},-\quad-x->-2--(a)-Show-that--f'(x)-=-\frac{-9}{(x---2)^2}---(3)--Given-that-P-is-a-point-on-C-such-that-f'(x)-=--1,--(b)-find-the-coordinates-of-P-Edexcel-A-Level Maths Pure-Question 2-2014-Paper 5.png$

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} (3) Given that P is a point on C su... show full transcript

Worked Solution & Example Answer: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} (3) Given that P is a point on C such that f'(x) = -1, (b) find the coordinates of P - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 5

Step 1

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

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Answer

To find the derivative of ( f(x) = \frac{4x + 1}{x - 2} ), we will apply the quotient rule, which states:

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

where ( u = 4x + 1 ) and ( v = x - 2 ).

Calculating the derivatives of u and v:

( u' = 4 )
( v' = 1 )

Now, substituting into the quotient rule:

$f'(x) = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2}$

This expands to:

$f'(x) = \frac{4x - 8 - 4x - 1}{(x - 2)^2} = \frac{-9}{(x - 2)^2}$

Thus, we have shown that ( f'(x) = \frac{-9}{(x - 2)^2} ).

Step 2

find the coordinates of P.

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Answer

Given that ( f'(x) = -1 ), we can set the derivative equal to -1:

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

To solve for x, first multiply both sides by ( (x - 2)^2 ):

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

This simplifies to:

$9 = (x - 2)^2$

Taking the square root of both sides:

$\sqrt{9} = x - 2 \implies x - 2 = 3 \text{ or } x - 2 = -3$

Thus, we find:

( x = 5 )
( x = -1 ) (not valid since ( x > 2 ))

Now, substituting ( x = 5 ) back into the original function to find the y-coordinate:

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

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

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} (3) Given that P is a point on C such that f'(x) = -1, (b) find the coordinates of P - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 5

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

find the coordinates of P.

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