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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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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.