Section 2.9 : Continuity
14. Use the Intermediate Value Theorem to show that \({w^2} - 4\ln \left( {5w + 2} \right) = 0\) has at least one root in the interval \(\left[ {0,4} \right]\). Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval,
First, you need to determine the value of “\(M\)” that you need to use and then actually use the Intermediate Value Theorem. So, go back to the IVT and compare the conclusions of the theorem and it should be pretty obvious what the \(M\) should be and then just check that the hypothesis (i.e. the “requirements” of the theorem) are met and you’ll pretty much be done.
Okay, let’s start off by defining,
\[f\left( w \right) = {w^2} - 4\ln \left( {5w + 2} \right)\hspace{0.5in}\& \,\,\hspace{0.5in}\,\,M = 0\]The problem is then asking us to show that there is a \(c\) in \(\left[ { 0,4} \right]\) so that,
\[f\left( c \right) = 0 = M\]but this is exactly the second conclusion of the Intermediate Value Theorem. So, let’s see that the “requirements” of the theorem are met.
First, the function is a sum of a polynomial (which is continuous everywhere) and a natural logarithm (which is continuous on \(w > - {\textstyle{2 \over 5}}\) - i.e. where the argument is positive) and so is continuous on the interval \(\left[ {0,4} \right]\). Note that this IS a requirement that MUST be met in order to use the IVT and it is the one requirement that is most often overlooked. If we don’t have a continuous function the IVT simply can’t be used.
Now all that we need to do is verify that \(M\) is between the function values as the endpoints of the interval. So,
\[f\left( 0 \right) = - 2.7726\hspace{0.5in}\hspace{0.25in}f\left( 4 \right) = 3.6358\]Therefore, we have,
\[f\left( 0 \right) = - 2.7726 < 0 < 3.6358 = f\left( 4 \right)\]So, by the Intermediate Value Theorem there must be a number \(c\) such that,
\[0 < c < 4\hspace{0.5in}\& \hspace{0.5in}f\left( c \right) = 0\]and we have shown what we were asked to show.