For real numbers $x > 1,$ find the minimum value of
\[\frac{x + 8}{\sqrt{x - 1}}.\]
Let $y = \sqrt{x - 1}.$  Then $y^2 = x - 1,$ so $x = y^2 + 1.$  Then
\[\frac{x + 8}{\sqrt{x - 1}} = \frac{y^2 + 9}{y} = y + \frac{9}{y}.\]By AM-GM,
\[y + \frac{9}{y} \ge 6.\]Equality occurs when $y = 3,$ or $x = 10,$ so the minimum value is $\boxed{6}.$
6
Level 3
Intermediate Algebra
Problem: For real numbers $x > 1,$ find the minimum value of
\[\frac{x + 8}{\sqrt{x - 1}}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the expression, we start by rewriting it as a function of \( x \): 

Let \( f(x) = \frac{x + 8}{\sqrt{x - 1}} \). 

Next, we need to determine the domain of \( f(x) \). Since \( x > 1 \), the expression \( \sqrt{x - 1} \) is defined and positive. 

To find the minimum value, we will take the derivative of \( f(x) \) and set it to zero. We apply the quotient rule for differentiation:

Let \( u = x + 8 \) and \( v = \sqrt{x - 1} \). Then, the derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Calculating \( u' = 1 \) and \( v' = \frac{1}{2\sqrt{x - 1}} \):

Now substituting these into the derivative:

\[ f'(x) = \frac{(1)(\sqrt{x - 1}) - (x + 8)\left(\frac{1}{2\sqrt{x - 1}}\right)}{x - 1} \]

Simplifying the numerator:

\[ \sqrt{x - 1} - \frac{x + 8}{2\sqrt{x - 1}} = \frac{2(x - 1) - (x + 8)}{2\sqrt{x - 1}} = \frac{x - 10}{2\sqrt{x - 1}} \]

Setting the numerator equal to zero to find critical points:

\[ x - 10 = 0 \Rightarrow x = 10 \]

Next, we check if this critical point is a minimum by evaluating the second derivative or using the first derivative test. 

Calculating \( f(10) \):

\[ f(10) = \frac{10 + 8}{\sqrt{10 - 1}} = \frac{18}{3} = 6 \]

To confirm that this is indeed a minimum, we can evaluate \( f(x) \) as \( x \) approaches 1 and as \( x \) approaches infinity:

As \( x \to 1^+ \), \( f(x) \to \infty \).  
As \( x \to \infty \), \( f(x) \to \infty \).

Thus, \( f(x) \) has a minimum at \( x = 10 \) and the minimum value is \( 6 \). 

Therefore, the final answer is \( 6 \).<|END_OF_LATENT|> The final answer is $6$.