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(D) Given $f(x) = x^2 + 3x - 3$. Completing the square, we get $f(x) = (x + \frac{3}{2})^2 - \frac{9}{4} - 3 = (x + \frac{3}{2})^2 - \frac{21}{4}$. Fo

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$\frac{1}{5}$

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(D) Given $f(x) = x^2 + 3x - 3$. Completing the square, we get $f(x) = (x + \frac{3}{2})^2 - \frac{9}{4} - 3 = (x + \frac{3}{2})^2 - \frac{21}{4}$. For the inverse to exist, the function must be monotonic. The vertex of the parabola is at $x = -\frac{3}{2}$, so the function is strictly increasing on $[-\frac{3}{2}, \infty)$. Thus, $\alpha = -\frac{3}{2}$. Let $y = (x + \frac{3}{2})^2 - \frac{21}{4}$. Then $x + \frac{3}{2} = \sqrt{y + \frac{21}{4}}$, so $g(x) = f^{-1}(x) = \sqrt{x + \frac{21}{4}} - \frac{3}{2}$. We need to find $g'(x)$ at $x = \alpha + \frac{5}{2} = -\frac{3}{2} + \frac{5}{2} = 1$. $g'(x) = \frac{d}{dx}(\sqrt{x + \frac{21}{4}} - \frac{3}{2}) = \frac{1}{2\sqrt{x + \frac{21}{4}}}$. At $x = 1$, $g'(1) = \frac{1}{2\sqrt{1 + \frac{21}{4}}} = \frac{1}{2\sqrt{\frac{25}{4}}} = \frac{1}{2 	imes \frac{5}{2}} = \frac{1}{5}$.

If $\alpha$ is the minimum value for which the inverse of $f(x)=x^2+3x-3$ exists in $[\alpha, \infty)$ and $g$ is the inverse of $f$, then find the value of $\frac{dg}{dx}$ at $x=\alpha+\frac{5}{2}$.

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