Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the \( x \)-values at which they occur. \( f(x)=x^{2}-10 x-9 ;[3,6] \) The absolute maximum value is \( \square \) at \( x=\square \) ? (Use a comma to separate answers as needed)

To find the absolute maximum and minimum values of the function \( f(x) = x^2 - 10x - 9 \) over the interval \( [3, 6] \), we first calculate the derivative: \[ f'(x) = 2x - 10. \] Next, we set the derivative equal to zero to find critical points: \[ 2x - 10 = 0 \] \[ x = 5. \] Now, we evaluate \( f(x) \) at the critical point and at the endpoints of the interval: 1. \( f(3) = 3^2 - 10(3) - 9 = 9 - 30 - 9 = -30. \) 2. \( f(5) = 5^2 - 10(5) - 9 = 25 - 50 - 9 = -34. \) 3. \( f(6) = 6^2 - 10(6) - 9 = 36 - 60 - 9 = -33. \) Now, we compare these values: - \( f(3) = -30 \) - \( f(5) = -34 \) - \( f(6) = -33 \) The absolute maximum value is \( -30 \) at \( x = 3 \), and the absolute minimum value is \( -34 \) at \( x = 5 \). So, the answer is: The absolute maximum value is \( -30 \) at \( x=3 \), and the absolute minimum value is \( -34 \) at \( x=5 \).