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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 Solution

First, 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 2

Note 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.