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The curve C has equation y = f(x) where f(x) = \frac{4x + 1}{x - 2}, \, 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}, \, 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}, \, 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}
Answer
To find the derivative ( f'(x) ) of the function ( f(x) = \frac{4x + 1}{x - 2} ), we will use the quotient rule, which states:
[ f'(x) = \frac{v u' - u v'}{v^2} ]
where ( u = 4x + 1 ) and ( v = x - 2 ). Calculating the derivatives:
- ( u' = 4 )
- ( v' = 1 )
Now, applying the quotient rule:
[ f'(x) = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} ]
Expanding the numerator:
[ = \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.
Answer
We know that ( f'(x) = -1 ). Setting the expression for ( f'(x) ) equal to -1:
[ \frac{-9}{(x - 2)^2} = -1 ]
Multiplying both sides by ( (x - 2)^2 ) gives:
[ -9 = -1 \cdot (x - 2)^2 ]
This simplifies to:
[ (x - 2)^2 = 9 ]
Taking the square root of both sides results in:
[ x - 2 = 3 \quad \text{or} \quad x - 2 = -3 ]
Thus, solving for ( x ) yields:
[ x = 5 \quad \text{or} \quad x = -1 ]
Since ( x > 2 ), we take ( x = 5 ). Now, substituting 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 ]
The coordinates of point P are therefore ( (5, 7) ).