I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 6.1 : Average Function Value
3. Find \({f_{{\rm{avg}}}}\) for \(f\left( x \right) = 4{x^2} - x + 5\) on \(\left[ { - 2,3} \right]\) and determine the value(s) of \(c\) in \(\left[ { - 2,3} \right]\) for which \(f\left( c \right) = {f_{{\rm{avg}}}}\).
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Start SolutionFirst, we need to use the formula for the notes in this section to find \({f_{{\rm{avg}}}}\).
\[{f_{{\rm{avg}}}} = \frac{1}{{3 - \left( { - 2} \right)}}\int_{{ - \,2}}^{3}{{4{x^2} - x + 5\,dx}} = \frac{1}{5}\left. {\left( {\frac{4}{3}{x^3} - \frac{1}{2}{x^2} + 5x} \right)} \right|_{ - 2}^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{83}}{6}}}\] Show Step 2Note that for the second part of this problem we are really just asking to find the value of \(c\) that satisfies the Mean Value Theorem for Integrals.
There really isn’t much to do here other than solve \(f\left( c \right) = {f_{{\rm{avg}}}}\).
\[\begin{align*}4{c^2} - c + 5 & = \frac{{83}}{6}\\ 4{c^2} - c - \frac{{53}}{6} & = 0 \hspace{0.5in} \Rightarrow \hspace{0.5in}c = \frac{{1 \pm \sqrt {1 - 4\left( 4 \right)\left( { - \frac{{53}}{6}} \right)} }}{{2\left( 4 \right)}} = \frac{{1 \pm \sqrt {\frac{{427}}{3}} }}{{2\left( 4 \right)}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 1.3663,\,\,\,1.6163}}\end{align*}\]So, unlike the example from the notes both of the numbers that we found here are in the interval and so are both included in the answer.