When is one of these functions greater than the other?

We want to find $x$ so that $$\begin{align*} & x(x-1)\ge (x-1)(3x-5) \\ \iff & x^2-x \ge 3x^2-8x+5 \\ \iff & 2x^2-7x+5\le 0 \\ \iff & (2x-5)(x-1)\le 0. \end{align*}$$

By considering the graph of $y=(2x-5)(x-1)$, we see that the solution is $1\le x\le \tfrac{5}{2}.$

Using our working from above, $f(x)-g(x)=-(2x-5)(x-1)$, so the graph of $y = f(x) - g(x)$ looks like this:

Since $S$ is the set of $x$ such that $f(x)-g(x)\ge 0$, the minimum value of $f(x)-g(x)$ on $S$ must be zero, when $x = 1$ or $\dfrac{5}{2}$.

To find the maximum, we can complete the square: $$\begin {align*} -2x^2+7x-5 &= -2\left(x-\frac{7}{4}\right)^2+\frac{49}{8}-5\\ &=-2\left(x-\frac{7}{4}\right)^2+\frac{9}{8}. \end {align*}$$

Since $2\left(x-\frac{7}{4}\right)^2 \ge 0$, we have $-2x^2+7x-5\le \tfrac{9}{8}$, so the maximum value of $f(x)-g(x)$ on $S$ is $\tfrac{9}{8}$, and this is attained at $x=\frac{7}{4}$. So we can sketch $f(x)-g(x)$ on $S$.