= Solution
{c}
Use $f,g$ from <4/b/i>. Their <inner products> with $x^2$ are
$$
\langle f,x^2\rangle=\frac23,\qquad
\langle g,x^2\rangle=\sqrt{\frac32}\,\frac{-1+0+1}{3}=0.
$$
Thus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, so
$$
\|h\|^2=\frac13\left(\frac19+\frac49+\frac19\right)=\frac29.
$$
By <4/b/ii> the three <polynomials> are <orthogonal vectors>, and normalizing $h$ gives
$$
\boxed{\mathcal B=\left(1,\sqrt{\frac32}\,x,\frac{3x^2-2}{\sqrt2}\right).}
$$
This <orthonormal set> is a <basis> of the three-dimensional <vector space> $V$. It is the result of the <Gram-Schmidt process> applied to $1,x,x^2$.
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