Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the \( x \)-values at which they occur. \( f(x)=x^{2}-4 x-3 ;[1,5] \) 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 \( f(x) = x^2 - 4x - 3 \) over the interval \([1, 5]\), we need to evaluate the function at the endpoints and any critical points within the interval. 1. **Endpoints Evaluation**: - At \( x = 1 \): \[ f(1) = 1^2 - 4(1) - 3 = 1 - 4 - 3 = -6 \] - At \( x = 5 \): \[ f(5) = 5^2 - 4(5) - 3 = 25 - 20 - 3 = 2 \] 2. **Finding Critical Points**: - To find critical points, we take the derivative \( f'(x) = 2x - 4 \) and set it to zero: \[ 2x - 4 = 0 \implies x = 2 \] - Now, we evaluate \( f \) at the critical point \( x = 2 \): \[ f(2) = 2^2 - 4(2) - 3 = 4 - 8 - 3 = -7 \] 3. **Comparison**: Now we compare the values found: - \( f(1) = -6 \) - \( f(2) = -7 \) - \( f(5) = 2 \) From these evaluations, the absolute maximum value is \( 2 \) at \( x = 5 \), and the absolute minimum value is \( -7 \) at \( x = 2 \). So the answers are: The absolute maximum value is \( 2 \) at \( x=5 \). The absolute minimum value is \( -7 \) at \( x=2 \).