Question

Find the critical numbers of the function.

\(p\left( x \right) = \frac{{{x^{\bf{2}}} + {\bf{2}}}}{{{\bf{2}}x - {\bf{1}}}}\)

Short answer

The critical numbers are \(x = - 1{\rm{ and }}2\).

Step-by-step solution

Critical Numbers of a function

TheCritical numbersfor any function \(p\left( x \right)\) are obtained by putting \(p'\left( x \right) = 0\).

Differentiating the function for critical numbers:

The given function is:

\(p\left( x \right) = \frac{{{x^2} + 2}}{{2x - 1}}\)

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

\(\begin{array}{c}p'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{{x^2} + 2}}{{2x - 1}}} \right)\\ = \frac{{\left( {2x - 1} \right)\frac{d}{{dx}}\left( {{x^2} + 2} \right) - \left( {{x^2} + 2} \right)\frac{d}{{dx}}\left( {2x - 1} \right)}}{{{{\left( {2x - 1} \right)}^2}}}\\ = \frac{{4{x^2} - 2x - 2{x^2} - 4}}{{{{\left( {2x - 1} \right)}^2}}}\\ = \frac{{2{x^2} - 2x - 4}}{{{{\left( {2x - 1} \right)}^2}}}\end{array}\)

Solving for critical numbers:

\(\begin{array}{c}p'\left( x \right) = 0\\\frac{{2{x^2} - 2x - 4}}{{{{\left( {2x - 1} \right)}^2}}} = 0\\{x^2} - x - 2 = 0\\x = - 1,2\end{array}\)

Here, the value \(\frac{1}{2}\) is not in the domain of \(p\), so the critical numbers are \(x = - 1{\rm{ and }}2\).