Answer
Average value = $\displaystyle \frac{16}{3}$
$x=\sqrt{3}$
Work Step by Step
If $f$ is integrable on the closed interval $[a, b]$, then the average value of $f$ on the interval is
$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$.
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Average value = $\displaystyle \frac{1}{3-1}\int_{1}^{3}\frac{4(x^{2}+1)}{x^{2}}dx$
$=2\displaystyle \int_{1}^{3}(1+x^{-2})dx$
$=2[x-\displaystyle \frac{1}{x}]_{1}^{3}$
$=2[(3-\displaystyle \frac{1}{3})-(1-\frac{1}{1})]$
$=\displaystyle \frac{16}{3}$
To find x where f(x)=6, solve $\displaystyle \frac{4(x^{2}+1)}{x^{2}}=\frac{16}{3}$ for x (on [1,3])
...Multiply both sides by $3x^{2}$
$12x^{2}+12=16x^{2}$
$12=16x^{2}-12x^{2}$
$4x^{2}=12$
$x^{2}=3$
$x=\pm\sqrt{3}$
On [1,3], $x=\sqrt{3}$
