= Solution
Expand $x=\sum_j c_jv_j$ in the <orthonormal basis> from part (a). Then
$$
\frac{x^\dagger Ax}{x^\dagger x}=\frac{\sum_j\lambda_j|c_j|^2}{\sum_j|c_j|^2}.
$$
For $x\ne0$, the weights $|c_j|^2/\sum_k|c_k|^2$ are nonnegative and sum to one. The <Rayleigh quotient> therefore lies between the smallest and largest <eigenvalues>. Conversely, normalized extreme <eigenvectors> $v_{\min},v_{\max}$ give, for $0\le t\le1$,
$$
x=\sqrt{1-t}\,v_{\min}+\sqrt t\,v_{\max},\qquad \frac{x^\dagger Ax}{x^\dagger x}=(1-t)\lambda_{\min}+t\lambda_{\max}.
$$
Thus \b[the exact <Hermitian Rayleigh quotient range> is]
$$
\boxed{[\lambda_{\min},\lambda_{\max}].}
$$
When $n=1$, the range is the singleton containing the sole <eigenvalue>.
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