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Section 4-2 : Critical Points

17. Determine the critical points of \(A\left( t \right) = 3t - 7\ln \left( {8t + 2} \right)\).

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Start Solution

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

\[A'\left( t \right) = 3 - 7\left( {\frac{8}{{8t + 2}}} \right) = 3 - \frac{{56}}{{8t + 2}} = \frac{{24t - 50}}{{8t + 2}}\]

We did quite a bit of simplification of the derivative to help with the next step. While not technically required it will mean the next step will be a fair amount simpler to do.

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 polyomial that is inside the logarithm and so any values of \(t\) for which the denominator is zero (i.e. \(t = - {\textstyle{1 \over 4}}\) since it’s easy to see that point) will not be in the domain of the original function (i.e. the function, \(A\left( { - {\textstyle{1 \over 4}}} \right)\), won’t exist because we can’t take the logarithm of zero). Therefore, this point will not be a critical point.

So, setting the numerator equal to zero gives,

\[24t - 50 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = {\textstyle{{25} \over {12}}}\] Show Step 3

As a final step we really should check that \(A\left( {{\textstyle{{25} \over {12}}}} \right)\) exists since there is always a chance that it won’t since we are dealing with a logarithm. It does exist (\(A\left( {{\textstyle{{25} \over {12}}}} \right) = {\textstyle{{75} \over {12}}} - 7\ln \left( {{\textstyle{{65} \over 3}}} \right)\)) and so the only critical point for this function is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = {\textstyle{{25} \over {12}}}}}\]