Question

Find the critical numbers of the function.

36. \(h\left( p \right) = \frac{{p - {\bf{1}}}}{{{p^{\bf{2}}} + {\bf{4}}}}\)

Short answer

The critical numbers are \(p = 1 \pm \sqrt 5 \).

Step-by-step solution

Critical Numbers of a function

TheCritical numbers for any function \(h\left( p \right)\) are obtained by putting \(h'\left( p \right) = 0\).

Differentiating the function for critical numbers:

The given function is:

\(h\left( p \right) = \frac{{p - 1}}{{{p^2} + 4}}\)

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

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

Solving for critical numbers:

\(\begin{array}{c}h'\left( p \right) = 0\\\frac{{ - {p^2} + 2p + 4}}{{{{\left( {{p^2} + 4} \right)}^2}}} = 0\\ - {p^2} + 2p + 4 = 0\\p = 1 \pm \sqrt 5 \end{array}\)

Hence, the critical numbers are \(p = 1 \pm \sqrt 5 \).