It is given that $f''(x) = e^{-\tfrac{1}{2}x} + \dfrac{5}{(5x-9)^2}$. Given that the x-axis is a tangent to the curve $y = f(x)$ at the point where $x = 2$,

1. Integrate $f''(x)$ to get the general form of $f'(x)$.

2. Use the tangent condition to find the constant of integration.

Since the $x$-axis is tangent to the curve at $x=2$, the gradient at that point is $0$.

So $f'(2)=0$.

1. Integrate $f'(x)$ to obtain the general form of $f(x)$.

From part (i),
$f'(x)=-2e^{-\frac12x}-\frac{1}{5x-9}+1+\frac2e.$
Integrating,
$f(x)=4e^{-\frac12x}-\frac15\ln(5x-9)+\left(1+\frac2e\right)x+C.$

2. Use the tangent-to-x-axis condition to find the constant.

Since the x-axis is tangent to the curve at $x=2$, the point of contact lies on the x-axis. Hence
$f(2)=0.$
Substitute $x=2$:
$0=4e^{-1}-\frac15\ln(1)+\left(1+\frac2e\right)(2)+C.$
Since $\ln(1)=0$,
$0=\frac4e+2+\frac4e+C$
$C=-2-\frac8e.$

1. Relate the gradient function to the second derivative.

The gradient of the curve $y=f(x)$ is $f'(x)$. To decide whether this gradient is increasing or decreasing, consider $f''(x).$

2. Determine the sign of $f''(x)$ for the given interval and conclude.

For $x>\dfrac95$, we have $5x-9>0$, so $(5x-9)^2>0$. Also, $e^{-\frac12x}>0.$ Hence $f''(x)=e^{-\frac12x}+\frac{5}{(5x-9)^2}>0.$ Therefore $f'(x)$ is increasing, so the gradient of the curve is an increasing function.