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Section 4.2 : Critical Points

16. Determine the critical points of \(R\left( x \right) = \ln \left( {{x^2} + 4x + 14} \right)\).

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

\[R'\left( x \right) = \frac{{2x + 4}}{{{x^2} + 4x + 14}}\] Show Step 2

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.

So, 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. However, in this case note that the denominator is also the polynomial that is inside the logarithm and so any values of \(x\) for which the denominator is zero will not be in the domain of the original function (i.e. the function, \(R\left( x \right)\), won’t exist at those values of \(x\) because we can’t take the logarithm of zero). Therefore, these points will not be critical points and we don’t need to bother determining where the derivative will be zero.

So, setting the numerator equal to zero gives,

\[2x + 4 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 2\] Show Step 3

As a final step we really should check that \(R\left( { - 2} \right)\) exists since there is always a chance that it won’t since we are dealing with a logarithm. It does exist (\(R\left( { - 2} \right) = \ln \left( {10} \right)\)) and so the only critical point for this function is,

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