Problem 119 - College Algebra

Question

Define a piecewise function on the intervals \((-\infty, 2],(2,5)\) and \([5, \infty)\) that does not "jump" at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

Step-by-Step Solution

Answer

The piecewise function that serves our purpose is: \(f(x) = 0\) for all x in \((- \infty, 2]\), \(f(x) = x - 2\) for all x in \((2, 5)\), and \(f(x) = 8 - x\) for all x in \([5, \infty)\).

Step 1 - Define Constant Function

For a function to be constant, it must remain the same throughout the interval. So, for the interval \((- \infty, 2]\), let's define the function as \(f(x) = c\), where c is any constant. For simplicity, let's choose c as 0. So, \(f(x) = 0\) for all x in \((- \infty, 2]\).

Step 2 - Define Increasing Function

Next, we need an increasing function on the interval \((2, 5)\). A simple example of an increasing function is a linear function with a positive slope, such as \(f(x) = x\). However, in order for our piecewise function to be continuous, we need to ensure that this function also equals 0 at \(x = 2\). We can accomplish this by defining our function as \(f(x) = x - 2\) for all x in \((2, 5)\). This function is increasing, and is equal to 0 at \(x = 2\), connecting smoothly with our previous piece.

Step 3 - Define Decreasing Function

Finally, we need a decreasing function on the interval \([5, \infty)\). Again, a linear function, now with a negative slope, can serve as a simple decreasing function. But, just like before, we need to ensure that this function equals 3 at \(x = 5\) for continuity. Therefore, we can define this function as \(f(x) = 8 - x\) for all x in \([5, \infty)\). This function is decreasing and equals 3 at \(x = 5\), so it connects smoothly with our previous piece.