Answer
Average value = 6
$x=\pm\sqrt{3}$
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 = $\displaystyle \frac{1}{3-(-3)}\int_{-3}^{3}(9-x^{2})dx$
$=\displaystyle \frac{1}{6}[9x-\frac{1}{3}x^{3}]_{-3}^{3}$
$=\displaystyle \frac{1}{6}[(27-9)-(-27+9)]$
$=6$
To find x where f(x)=6, solve $9-x^{2}=6$ for x
$x^{2}=9-6$
$x^{2}=3$
$x=\pm\sqrt{3}$
