Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the \( x \)-values at which they occur: \( f(x)=8 x-4 \) \( \begin{array}{ll}\text { (A) }[0,5] & \text { (B) }[-6,5] \\ \text { (A) Find the first derivative of } f\end{array} \) \( f^{\prime}(x)=\square \)

To find the absolute maximum and minimum of the function \( f(x) = 8x - 4 \) over the given intervals, let's start by finding the first derivative of the function. The first derivative is: \[ f'(x) = 8. \] Since this derivative is constant, it means that the function is linear and does not change direction—it continuously increases throughout the interval. Next, we evaluate the function at the endpoints of the intervals to find the absolute maximum and minimum values. For interval (A) \([0, 5]\): - \( f(0) = 8(0) - 4 = -4 \) - \( f(5) = 8(5) - 4 = 36 \) For interval (B) \([-6, 5]\): - \( f(-6) = 8(-6) - 4 = -52 \) - \( f(5) = 8(5) - 4 = 36 \) Now we can summarize the results: - In interval (A), the absolute minimum value is \(-4\) at \(x = 0\) and the absolute maximum value is \(36\) at \(x = 5\). - In interval (B), the absolute minimum value is \(-52\) at \(x = -6\) and the absolute maximum value remains \(36\) at \(x = 5\). So for your intervals: - (A) Absolute Minimum: -4 at \(x=0\); Absolute Maximum: 36 at \(x=5\). - (B) Absolute Minimum: -52 at \(x=-6\); Absolute Maximum: 36 at \(x=5\).