1. The problem asks us to find the greatest value of the expression $$\frac{1}{4x^2 - 4x + 7}$$ and the value of $x$ at which this maximum occurs. 2. From part (a), we have rewritten the quadratic expression in the denominator as $$4x^2 - 4x + 7 = 4(x - \frac{1}{2})^2 + 6$$ where $p=4$, $q=-\frac{1}{2}$, and $r=6$. 3. Since the denominator is $4(x - \frac{1}{2})^2 + 6$, and squares are always non-negative, the smallest value of the denominator occurs when the squared term is zero, i.e., when $$x - \frac{1}{2} = 0 \implies x = \frac{1}{2}$$. 4. At this $x$ value, the denominator is minimized to $$4 \times 0^2 + 6 = 6$$. 5. Because the expression is $$\frac{1}{\text{denominator}}$$, the greatest value of the whole expression occurs when the denominator is smallest. 6. Therefore, the greatest value is $$\frac{1}{6}$$, and it occurs at $$x = \frac{1}{2}$$. 7. We assume the minimum of the denominator because the reciprocal function $$f(y) = \frac{1}{y}$$ is decreasing for positive $y$, so minimizing the denominator maximizes the fraction. This explains why we find the minimum of the quadratic in the denominator to get the maximum of the reciprocal function.