Question

Find the critical numbers of the function.

38. \(q\left( t \right) = \frac{{{t^{\bf{2}}} + {\bf{9}}}}{{{t^{\bf{2}}} - {\bf{9}}}}\)

Short answer

The only critical number is \(t = 0\).

Step-by-step solution

Critical Numbers of a function

TheCritical numbersfor any function \(q\left( t \right)\) are obtained by putting \(q'\left( t \right) = 0\).

Differentiating the function for critical numbers:

The given function is:

\(q\left( t \right) = \frac{{{t^2} + 9}}{{{t^2} - 9}}\)

On differentiating with respect to\(x\)as:

\(\begin{array}{c}q'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{{t^2} + 9}}{{{t^2} - 9}}} \right)\\ = \frac{{\left( {{t^2} - 9} \right)\frac{d}{{dx}}\left( {{t^2} + 9} \right) - \left( {{t^2} + 9} \right)\frac{d}{{dx}}\left( {{t^2} - 9} \right)}}{{{{\left( {{t^2} - 9} \right)}^2}}}\\ = \frac{{2{t^3} - 18t - 2{t^3} - 18t}}{{{{\left( {{t^2} - 9} \right)}^2}}}\\ = \frac{{ - 36t}}{{{{\left( {{t^2} - 9} \right)}^2}}}\end{array}\)

Solving for critical numbers:

\(\begin{array}{c}q'\left( t \right) = 0\\\frac{{ - 36t}}{{{{\left( {{t^2} - 9} \right)}^2}}} = 0\\ - 36t = 0\\t = 0\end{array}\)

Here, \(t \ne 3, - 3\) because it is not in the domain of \(q\), hence, the critical number obtained is \(t = 0\).