Section 4.2 : Critical Points
8. Determine the critical points of \(\displaystyle R\left( x \right) = \frac{{1 - x}}{{{x^2} + 2x - 15}}\).
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Start SolutionWe’ll need the first derivative to get the answer to this problem so let’s get that.
\[R'\left( x \right) = \frac{{\left( { - 1} \right)\left( {{x^2} + 2x - 15} \right) - \left( {1 - x} \right)\left( {2x + 2} \right)}}{{{{\left( {{x^2} + 2x - 15} \right)}^2}}} = \frac{{{x^2} - 2x + 13}}{{{{\left( {{x^2} + 2x - 15} \right)}^2}}}\] Show Step 2Recall 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 that the exponent on the whole denominator will not affect where it is zero and so can be ignored. This means we need to solve the following two equations.
\[{x^2} - 2x + 13 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{{2 \pm \sqrt { - 48} }}{2} = 1 \pm 2\sqrt 3 \,i\] \[{x^2} + 2x - 15 = \left( {x + 5} \right)\left( {x - 3} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 5,\,\,3\]As we can see in this case we needed to use the quadratic formula on the first quadratic equation. Remember that not all quadratics will factor so don’t forget about the quadratic formula!
Show Step 3Now, recall that we don’t use complex numbers in this class and so the solutions from where the numerator is zero won’t be critical points.
Also recall that a point will only be a critical point if the function (not the derivative, but the original function) exists at the point. For this problem we found two values where the derivative doesn’t exist, however the function also doesn’t exist at these points and so neither of these will be critical points either.
Therefore, this function has no critical points. Do not get excited about this when it happens. Not all functions will have critical points!