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Section 4-2 : Critical Points

7. Determine the critical points of \(\displaystyle f\left( z \right) = \frac{{z + 4}}{{2{z^2} + z + 8}}\).

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We’ll need the first derivative to get the answer to this problem so let’s get that.

\[f'\left( z \right) = \frac{{\left( 1 \right)\left( {2{z^2} + z + 8} \right) - \left( {z + 4} \right)\left( {4z + 1} \right)}}{{{{\left( {2{z^2} + z + 8} \right)}^2}}} = \frac{{ - 2{z^2} - 16z + 4}}{{{{\left( {2{z^2} + z + 8} \right)}^2}}} = \frac{{ - 2\left( {{z^2} + 8z - 2} \right)}}{{{{\left( {2{z^2} + z + 8} \right)}^2}}}\]

The “-2” was factored out of the numerator only to make it a little nicer for the next step and doesn’t really need to be done.

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Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. Therefore, we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero.

So, all we need to do is set the numerator and denominator equal to zero and solve. Note as well that the “-2” we factored out of the numerator will not affect where it is zero and so can be ignored. Likewise, the exponent on the whole denominator will not affect where it is zero and so can also be ignored. This means we need to solve the following two equations.

\[{z^2} + 8z - 2 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}z = \frac{{ - 8 \pm \sqrt {72} }}{2} = - 4 \pm 3\sqrt 2 \] \[2{z^2} + z + 8 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}z = \frac{{ - 1 \pm \sqrt { - 63} }}{4} = \frac{{ - 1 \pm \sqrt {63} \,i}}{4}\]

As we can see in this case we needed to use the quadratic formula both of the quadratic equations. Remember that not all quadratics will factor so don’t forget about the quadratic formula!

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Now, recall that we don’t use complex numbers in this class and so the solutions from where the denominator is zero (i.e. the derivative doesn’t exist) won’t be critical points. Therefore, the only critical points of this function are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 4 \pm 3\sqrt 2 }}\]