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Section 2.9 : Continuity
9. Determine where the following function is discontinuous.
\[R\left( t \right) = \frac{{8t}}{{{t^2} - 9t - 1}}\]
Hint : If we have two continuous functions and form a rational expression out of them recall where the rational expression will be discontinuous. We discussed this in the Limit Properties section, although we were using the phrase “nice enough” there instead of the word “continuity”.
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since both are polynomials) the only points in which the rational expression will be discontinuous will be where we have division by zero.
Therefore, all we need to do is determine where the denominator is zero and that is fairly easy for this problem.
\[{t^2} - 9t - 1 = 0\hspace{0.25in}\,\,\,\, \Rightarrow \,\hspace{0.5in}t = \frac{{9 \pm \sqrt {{{\left( { - 9} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} = \frac{{9 \pm \sqrt {85} }}{2} = - 0.10977,\,\,9.10977\]The function will therefore be discontinuous at the points : \(t = \frac{{9 \pm \sqrt {85} }}{2}\).