The curve C has equation $y = f(x), x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ Given that the point $P(4, -8)$ lies on C, (a) find the equation of the tangent to C at P, giving your answer in the form $y = mx + c$, where $m$ and $c$ are constants - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1

To find $f(x)$, we integrate $f'(x)$:

$f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$

1. Rewrite the second term:

   $\frac{6 - 5x^2}{\sqrt{x}} = \frac{6}{\sqrt{x}} - 5x^{\frac{3}{2}}$

   Therefore:

   $f'(x) = 30 + 6x^{-\frac{1}{2}} - 5x^{\frac{3}{2}}$

2. Now, integrate term by term:

   $f(x) = \int(30)dx + \int(6x^{-\frac{1}{2}})dx - \int(5x^{\frac{3}{2}})dx$

   This leads to:

   $f(x) = 30x + 6 \cdot 2x^{\frac{1}{2}} - 5\cdot \frac{2}{5} x^{\frac{5}{2}} + C$

   Simplifying gives:

   $f(x) = 30x + 12\sqrt{x} - 2x^{\frac{5}{2}} + C$

3. Using the point $P(4, -8)$ to find $C$:

   $-8 = 30(4) + 12\sqrt{4} - 2(4)^{\frac{5}{2}} + C$

   Calculating the terms gives:

   $$$$

ightarrow C = -88$$

Therefore,

$f(x) = 30x + 12\sqrt{x} - 2x^{\frac{5}{2}} - 88$