Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the \( x \)-values at which they occur. \( f(x)=x^{2}-2 x-3 ;[-3,4] \) The absolute maximum value is \( \square \) at \( x=\square \) (Use a comma to separate answers as needed.) The absolute minimum 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 \( f(x) = x^2 - 2x - 3 \) over the interval \([-3, 4]\), we need to evaluate the function at the critical points and endpoints. 1. **Find the critical points**: First, we take the derivative of the function: \[ f'(x) = 2x - 2 \] Setting the derivative equal to zero gives: \[ 2x - 2 = 0 \implies x = 1 \] Now we check if \( x = 1 \) is within the interval \([-3, 4]\). It is. 2. **Evaluate \( f(x) \) at the endpoints and the critical point**: - At \( x = -3 \): \[ f(-3) = (-3)^2 - 2(-3) - 3 = 9 + 6 - 3 = 12 \] - At \( x = 1 \): \[ f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4 \] - At \( x = 4 \): \[ f(4) = (4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5 \] 3. **Compare the values**: - \( f(-3) = 12 \) - \( f(1) = -4 \) - \( f(4) = 5 \) From this, we find that the absolute maximum value is \( 12 \) at \( x = -3 \) and the absolute minimum value is \( -4 \) at \( x = 1 \). Thus, the answers are: The absolute maximum value is \( 12 \) at \( x = -3 \) The absolute minimum value is \( -4 \) at \( x = 1 \)