Solution:

(a)  A student should immediately factorise the denominator to identify the vertical asymptotes for the graph of the function.

\(f\left( x \right) = \frac{1}{{{x^2} + 3x - 10}} = \frac{1}{{\left( {x + 5} \right)\left( {x - 2} \right)}}\)

Thus, the graph of f has vertical asymptotes (and discontinuities) at \(x = - 5\) and at \(x = 2\).  The domain is certain to be \(x \in \mathbb{R},\;x \ne - 5,\;x \ne 2\).  Determining the range is much less clear.  One can figure out that the graph of f will have a horizontal asymptote \(y = 0\) (x-axis) because as x becomes either a very large positive or very large negative number the \({x^2}\) term will dominate and cause the denominator \({x^2} + 3x - 10\) to become a large positive number.  Thus, as \(x \to + \,\infty \) or as \(x \to - \,\infty \) the value of \(\frac{1}{{{x^2} + 3x - 10}}\) approaches zero from positive values meaning that the graph of f will be asymptotic to the x-axis from above (i.e. the end behaviour of f).  So, an educated guess is that the range is \(y \in \mathbb{R},\;y \ne 0\).

Graphing on a GDC (see images below) confirms the domain as stated, but the ‘gap’ in y values is more than just zero.  The downward-opening branch of the graph of f appears to have a maximum value less than zero.  Analyzing the graph on the GDC reveals that this downward-opening branch of the graph has a maximum at \(\left( { - 1.5, - \,0.08163} \right)\).  It’s clear that the x-coordinate of this local maximum point is exactly \( - 1.5\) but what about the y-coordinate?  Finding it exactly requires some effort.  Somewhat surprisingly, it’s more of an effort on the TI-Nspire than the TI-84.

After converting to a fraction, it’s revealed that the exact value of the local maximum’s y-coordinate is \( - \frac{4}{{49}}\).  Therefore, the range of f is \(y \le - \frac{4}{{49}},\;y > 0\).  Note:  Although \( - \frac{4}{{49}} \approx - \,0.0816\), it is incorrect to state the range as \(y \le - 0.0816,\;y > 0\) or as \(y < - 0.0816,\;y > 0\).  The ‘endpoint’ of an inequality used for indicating the domain or range must be expressed exactly.  For example, the value \(y = - \,0.8162\) is less than \( - \,0.816\); but \( - \frac{4}{{49}} \approx - \,0.081632653061...\) and \( - \,0.8162\) is not less than \( - \,0.081632653061...\).  Hence, \(y = - \,0.8162\) is not in the range but expressing the range as either \(y \le - 0.0816,\;y > 0\) or \(y < - 0.0816,\;y > 0\) would incorrectly indicate that \(y = - \,0.8162\) is in the range.

(b)   For determining the domain of g we know that the graph of \(g\left( x \right) = \sqrt {\frac{{8x - 4}}{{x - 3}}} \) has a vertical asymptote (and discontinuity) at \(x = 3\).  It is also necessary to find where \(\frac{{8x - 4}}{{x - 3}} > 0\) which takes a bit of effort (remember this is for the purpose of making an ‘educated guess’ before looking at a graph of the function on a GDC).  A sign chart will work for solving the inequality.

So, it seems that the domain should be \(x < \frac{1}{2},\;x > 3\).

Working out the range initially seems less tricky than the previous example.  Taking the square root can only give non-negative results.  Hence, an educated guess for the range is \(y \ge 0\).

A graph of g confirms the conjecture for the domain but appears to indicate there is a horizontal asymptote with an equation somewhere around \(y = 3\). As with the function in part (a), considering the end behaviour of the function will lead to an exact equation for the horizontal asymptote.  When \(x \to + \,\infty \) or \(x \to - \,\infty \), the x terms in the numerator and denominator of \(\frac{{8x - 4}}{{x - 3}}\) will dominate so that the value of \(\frac{{8x - 4}}{{x - 3}}\) will come ever closer to \(\frac{8}{1}\).  Thus, the equation of the horizontal asymptote is \(y = \sqrt 8 = 2\sqrt 2 \); also indicating where there will be a ‘gap’ in the range.  Therefore, the correct range for function g is \(0 \le y < 2\sqrt 2 ,\;y > 2\sqrt 2 \).