Consider the function \( f \) given below. \( f(x)=\frac{x^{2}-81}{x-9} \) a) For what \( x \)-values(s) is this function not differentiable? b) Find \( f^{\prime}(-1) \).

To analyze the function \( f(x) = \frac{x^2 - 81}{x - 9} \), we can start by simplifying it. ### Step 1: Simplifying the Function The expression \( x^2 - 81 \) can be factored as follows: \[ x^2 - 81 = (x - 9)(x + 9) \] Thus, we can rewrite \( f(x) \): \[ f(x) = \frac{(x - 9)(x + 9)}{x - 9} \] For \( x \neq 9 \), we can simplify this to: \[ f(x) = x + 9 \] However, at \( x = 9 \), the original function is undefined because it results in a division by zero. ### Part a: Points of Non-Differentiability The function \( f(x) \) is not differentiable at points where it is not defined. Since \( f(x) \) is undefined at \( x = 9 \), this is the only point where the function is not differentiable. **Answer for part a:** The function is not differentiable at \( x = 9 \). ### Part b: Finding \( f'(-1) \) Since \( f(x) = x + 9 \) for \( x \neq 9 \), we can differentiate this simplified function: \[ f'(x) = 1 \] This derivative holds for all \( x \) except \( x = 9 \). Now, we can find \( f'(-1) \): \[ f'(-1) = 1 \] **Answer for part b:** \( f'(-1) = 1 \).