Problem 119 - College Algebra Essentials

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 which meets all the specified conditions is \(f(x) = 3\) for \(-\infty < x \leq 2\), \(f(x) = x + 1\) for \(2 < x < 5\), and \(f(x) = -x + 11\) for \(5 \leq x < \infty\).

Step 1: Define the Constant Function Segment

The first part of the problem can be a constant function, because such a function doesn't increase or decrease. Set the function equal to a constant over the first interval. For example, \(f(x) = c\) for \( -\infty < x \leq 2\), where \(c\) could be any real number. Let's choose \(c = 3\). So the first piece of the function \(f(x)\) can be \(f(x) = 3\) for \(-\infty < x \leq 2\).

Step 2: Define the Increasing Function Segment

For the increasing function on the interval (2,5), a simple choice is a linear function which increases as \(x\) increases, but it must also match the constant function at \(x = 2\) to maintain continuity. For simplicity, choose a function that slopes upwards like \(f(x) = x\). However, this function at \(x = 2\) would be \(2\), whereas the constant function gives \(3\) at \(x = 2\). To rectify this, the function \(f(x) = x + 1\) is chosen. This will give \(f(2) = 3\), the same as the constant function, and will increase afterwards. So, the second piece of the function \(f(x)\) can be \(f(x) = x + 1\) for \(2 < x < 5\).

Step 3: Define the Decreasing Function Segment

For the decreasing function, a linear function such as \(f(x) = -x\) could also work, but it would be -5 at \(x = 5\), whereas it needs to be 6 to match the increasing function. Thus, it's better to define the function \(f(x) = -x + 11\). So the third piece of the function \(f(x)\) can be \(f(x) = -x + 11\) for \(5 \leq x < \infty\).