For a function $f$, defined on a suitable domain, it is known that: • $f^{\prime}(x) = \frac{2x+1}{\sqrt{x}}$ • $f(9) = 40$ Express $f(x)$ in terms of $x$. - Scottish Highers Maths - Question 9 - 2016

Integrating term by term:

1. For $2x^{\frac{1}{2}}$:
   $\int 2x^{\frac{1}{2}} \, dx = \frac{2}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}}$

2. For $x^{-\frac{1}{2}}$:
   $\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}}$

Combining these results:

$f(x) = \frac{4}{3} x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + c$

Now we will use the condition $f(9) = 40$ to find the constant $c$:

Substituting $x = 9$:

$f(9) = \frac{4}{3} (9^{\frac{3}{2}}) + 2(9^{\frac{1}{2}}) + c$
Calculating the powers:
$9^{\frac{3}{2}} = 27, \quad 9^{\frac{1}{2}} = 3$
Thus,

$f(9) = \frac{4}{3} \times 27 + 2 \times 3 + c = 36 + 6 + c = 42 + c$ Setting it equal to 40:

$42 + c = 40$ Solving for $c$:
$c = 40 - 42 = -2$

The expression for $f(x)$ is therefore:

$f(x) = \frac{4}{3} x^{\frac{3}{2}} + 2x^{\frac{1}{2}} - 2$