Answer
see the proof below.
Work Step by Step
Let $f(x)= x, g(x)=\frac{1}{2}(3x^2-1) $, $C[-1,1]$
\begin{aligned}\langle f,g\rangle &=\int_{-1}^{1}\frac{1}{2}(3x^3-x) d x\\
&=\left[\frac{1}{2}(\frac{3}{4}x^4-\frac{1}{2}x^2) \right]_{-1}^{1} \\ &=0 \end{aligned}.
then $f$ and $g$ are orthogonal in the inner product space $C[-1,1]$.