Problem 31 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=5-|x-5| $$

Step-by-Step Solution

Answer

The function \(f(x)=5-|x-5|\) has a critical point at \(x = 5\). It increases on the interval \(x < 5\), and decreases on the interval \(x > 5\). The function also has a local maximum at \(x = 5\). This is confirmed by graphing the function.

Step 1: Finding the Critical Points

The critical points of a function occur where the derivative of the function is zero or undefined. However, for functions involving absolute values, we must consider the changes in direction separately. In this case, the absolute function \(|x-5|\) changes direction at \(x = 5\). Thus, \(x=5\) is a critical point.

Step 2: Determining the Intervals

The function's behavior changes at critical points. Therefore, we need to test the nature of function \(f\) for \(x < 5\) and \(x > 5\). If \(x < 5\), \(f(x) = 5 - (5 - x) = x\). If \(x > 5\), \(f(x) = 5 - (x - 5) = 10 - x\). Thus, the function is increasing for \(x < 5\) and decreasing for \(x > 5\).

Step 3: Determining the Extrema

A local maximum occurs when the function changes from increasing to decreasing. Since our function increases for \(x < 5\) and decreases for \(x > 5\), there is a local maximum at \(x = 5\). By plugging \(x = 5\) into \(f(x)\), we find that the maximum value is \(f(5) = 5\).

Step 4: Verifying the Results Graphically

By plotting the function graphically, we can verify that \(x = 5\) is a local maximum and the function increases and decreases where we have determined.