Answer
\[\begin{align}
& {{f}_{avg}}=\frac{1}{2} \\
& x=-\frac{1}{2}\text{ or }x=\frac{1}{2} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=1-\left| x \right|\text{ on }\left[ -1,1 \right] \\
& \text{By the definition of the absolute value of }g\left( x \right) \\
& g\left( x \right)=x\text{ for }x\ge 0 \\
& g\left( x \right)=-x\text{ for }x<0 \\
& \text{Then} \\
& f\left( x \right)=1-x,\text{ for }x\ge 0 \\
& f\left( x \right)=1+x,\text{ for }x<0 \\
& \text{Now calculating the average value of }f\left( x \right)\text{ on }\left[ -1,1 \right] \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{1-\left( -1 \right)}\int_{-1}^{1}{\left( 1-\left| x \right| \right)dx} \\
& {{f}_{avg}}=\frac{1}{2}\left[ \int_{-1}^{0}{\left( 1+x \right)dx}+\int_{0}^{1}{\left( 1-x \right)dx} \right] \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{2}\left( \left[ x+\frac{{{x}^{2}}}{2} \right]_{-1}^{0}+\left[ x-\frac{{{x}^{2}}}{2} \right]_{0}^{1} \right) \\
& {{f}_{avg}}=\frac{1}{2}\left( -\left[ \left( -1 \right)+\frac{{{\left( -1 \right)}^{2}}}{2} \right]+\left[ \left( 1 \right)-\frac{{{\left( 1 \right)}^{2}}}{2} \right] \right) \\
& \text{Simplifying} \\
& {{f}_{avg}}=\frac{1}{2}\left( \frac{1}{2}+\frac{1}{2} \right) \\
& {{f}_{avg}}=\frac{1}{2} \\
& \text{The point at which the given function equals its average value is} \\
& 1-\left| x \right|=\frac{1}{2} \\
& \text{Solve for }x \\
& \left| x \right|=1-\frac{1}{2} \\
& \left| x \right|=\frac{1}{2} \\
& x=-\frac{1}{2}\text{ or }x=\frac{1}{2} \\
\end{align}\]
