The curve C has equation y = f(x), x > 0, and f'(x) = 4x - 6rac{ ext{s}}{x} + rac{8}{x^2} - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

To find f(x), we start with the derivative given:

$f'(x) = 4x - 6 \frac{1}{x} + 8 \frac{1}{x^2}$

Next, we will integrate f'(x):

$f(x) = \int (4x - 6 \frac{1}{x} + 8 \frac{1}{x^2}) \; dx$

Calculating the integral term-by-term:

1. For the first term:
   $\int 4x \; dx = 2x^2$

2. For the second term:
   $\int -6 \frac{1}{x} \; dx = -6 \ln |x|$

3. For the third term:
   $\int 8 \frac{1}{x^2} \; dx = -\frac{8}{x}$

Thus, combining these results, we have:

$f(x) = 2x^2 - 6\ln |x| - \frac{8}{x} + C$

To find the constant C, we use the point P(4, 1) which lies on C:

Substituting x = 4 into the equation gives:

$1 = 2(4^2) - 6\ln |4| - \frac{8}{4} + C$

Calculating:

1. $2(16) = 32$
2. $-6\ln(4) \approx -6(1.386) \approx -8.316$
3. $-\frac{8}{4} = -2$

Substituting these values:

$1 = 32 - 8.316 - 2 + C$ $C = 1 - 32 + 8.316 + 2 \approx -20.684$

Therefore, the function simplifies to:

$f(x) = 2x^2 - 6\ln |x| - \frac{8}{x} - 20.684$