The curve C has equation $y=f(x)$, where $x > 0$, where $$\frac{dy}{dx} = \frac{3x^2 - \frac{5}{\sqrt{x}} - 2}{2}$$ Given that the point P(4, 5) lies on C, find (a) $f(x)$, (b) an equation of the tangent to C at the point P, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers. - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 1

To find $f(x)$, we need to integrate the expression for $\frac{dy}{dx}$.

1. We start with the differential equation: $\frac{dy}{dx} = \frac{3x^2 - \frac{5}{\sqrt{x}} - 2}{2}$

2. Simplifying the equation: $\frac{dy}{dx} = \frac{3x^2}{2} - \frac{5}{2\sqrt{x}} - 1$

3. Integrating each term in the expression: $y = \int \left(\frac{3x^2}{2} - \frac{5}{2\sqrt{x}} - 1\right) dx$

   • The integral of $\frac{3x^2}{2}$ is $\frac{3}{2} \cdot \frac{x^3}{3} = \frac{x^3}{2}$.
   • The integral of $-\frac{5}{2\sqrt{x}}$ is $-\frac{5}{2} \cdot 2\sqrt{x} = -5\sqrt{x}$.
   • The integral of $-1$ is $-x$.

4. Therefore, we have: $y = \frac{x^3}{2} - 5\sqrt{x} - x + C$

5. To find the constant $C$, we use the point P(4, 5): $5 = \frac{4^3}{2} - 5\sqrt{4} - 4 + C$ $5 = \frac{64}{2} - 10 - 4 + C$ $5 = 32 - 10 - 4 + C$ $5 = 18 + C$ $C = 5 - 18 = -13$

6. Thus, the function is: $f(x) = \frac{x^3}{2} - 5\sqrt{x} - x - 13$