Questions: Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. f(x)=(x^2-6x+9)/(x-5) Select the correct choice below and, if necessary, fill in any answer boxes within your choice. A. There are no relative maxima. The function has a relative minimum of at x= (Use a comma to separate answers as needed.). B. There are no relative minima. The function has a relative maximum of at x= (Use a comma to separate answers as needed.) C. The function has a relative maximum of at x= at x= and a relative minimum of (Use a comma to separate answers as needed.) D. There are no relative extrema.

Solution Steps

To find the relative extrema of the function \( f(x) = \frac{x^2 - 6x + 9}{x - 5} \), we need to follow these steps:

1. Find the derivative: Compute the derivative of the function using the quotient rule.
2. Critical points: Set the derivative equal to zero and solve for \( x \) to find critical points.
3. Second derivative test: Use the second derivative to determine the nature of each critical point (i.e., whether it is a relative maximum, minimum, or neither).

We start with the function

\[ f(x) = \frac{x^2 - 6x + 9}{x - 5}. \]

Using the quotient rule, we find the first derivative:

\[ f'(x) = \frac{(2x - 6)(x - 5) - (x^2 - 6x + 9)(1)}{(x - 5)^2}. \]

Setting the first derivative equal to zero, we solve for \( x \):

\[ f'(x) = 0 \implies (2x - 6)(x - 5) - (x^2 - 6x + 9) = 0. \]

This yields the critical points:

\[ x = 3, \quad x = 7. \]

Next, we compute the second derivative:

\[ f''(x) = 2/(x - 5) - 2(2x - 6)/(x - 5)^2 + 2(x^2 - 6x + 9)/(x - 5)^3. \]

Evaluating the second derivative at the critical points:

• For \( x = 3 \):

\[ f''(3) > 0 \implies \text{relative maximum}. \]

• For \( x = 7 \):

\[ f''(7) < 0 \implies \text{relative minimum}. \]

Final Answer