The point P(4, -1) lies on the curve C with equation y = f(x), x > 0, and $f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3$ (a) Find the equation of the tangent to C at the point P, giving your answer in the form y = mx + c, where m and c are integers - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 2

To find f(x), we integrate the derivative f'(x):

$f'(x) = \frac{1}{2} - \frac{6}{\sqrt{x}} + 3$

Integrating term by term results in:

$f(x) = \int \left( \frac{1}{2} - 6x^{-\frac{1}{2}} + 3 \right) dx$

This gives us:

$f(x) = \frac{1}{2}x - 12\sqrt{x} + 3x + c$

Now substituting the point P(4, -1) to find c:

$f(4) = \frac{1}{2}(4) - 12\sqrt{4} + 3(4) + c$

Calculating:

$-1 = 2 - 12(2) + 12 + c$

This simplifies to:

$-1 = 2 - 24 + 12 + c \Rightarrow -1 = -10 + c \Rightarrow c = 9$

Thus, the function is:

$f(x) = \frac{1}{2}x - 12\sqrt{x} + 3x + 9$

In expanded form:

$f(x) = \left(\frac{1}{2} + 3\right)x - 12\sqrt{x} + 9 = 3.5x - 12\sqrt{x} + 9$