(a)
Write:
\displaystyle f'(x) = \frac 3 2 x^{1/2} - \frac 9 4 x^{-1/2} + 2
Then:
\begin{align*}f(x) &= \int f'(x) \mathrm dx \\ & = \frac 3 2 \times \frac {x^{3/2}} {\frac 3 2} - \frac 9 4 \frac {x^{1/2}} {\frac 1 2} + 2x + C \\ & = x^{3/2} - \frac 9 2 x^{1/2} + 2x + C\end{align*}
At x = 4 we have:
\displaystyle y = 4^{3/2} - \frac 9 2 \times 4^{1/2} + 2 \times 4 = 8 - 9 + 8 + C = 7 + C = 9
So C = 2. Hence:
\displaystyle f(x) = x^{3/2} - \frac 9 2 x^{1/2} + 2x + 2
(b)
The line 2y + x = 0 has gradient -\dfrac 1 2. The normal at P therefore has gradient -\dfrac 1 2, so the tangent at P has gradient 2. So, we have:
\displaystyle \frac 3 2 \sqrt x - \frac 9 {4 \sqrt x} + 2 = 2
So:
\displaystyle \frac 3 2 \sqrt x = \frac 9 {4 \sqrt x}
So the x-coordinate of P is given by:
\displaystyle x = \frac 2 3 \times \frac 9 4 = \frac 3 2
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