Problem 71 - College Algebra with Modeling and Visualization

Question

Exercises \(69-74:\) Complete the following for \(f(x)\) (a) Determine the domain of \(f\) (b) Evaluate \(f(-2), f(0),\) and \(f(3)\) (c) Graph \(f\) (d) Is \(f\) continuous on its domain? $$ f(x)=\left\\{\begin{array}{ll} 3 x & \text { if }-1 \leq x<1 \\ x+1 & \text { if } \quad 1 \leq x \leq 2 \end{array}\right. $$

Step-by-Step Solution

Answer

(a) Domain is \([-1, 2]\). (b) \(f(-2)\) and \(f(3)\) are undefined; \(f(0) = 0\). (c) Piecewise lines for \(-1 \leq x < 1\) and \(1 \leq x \leq 2\). (d) \(f\) is not continuous.

Step 1: Determine the Domain

The domain of a function includes all possible values of \(x\) for which the function is defined. In this case, \(f(x)\) is a piecewise function with two conditions: \(-1 \leq x < 1\) and \(1 \leq x \leq 2\). The common domain for both these conditions is from \(-1\) to \(2\), but note that \(x=1\) is only covered by the second condition. Thus, the domain of \(f\) is \([-1, 2]\).

Step 2: Evaluate at Specific Points

We will evaluate \(f(x)\) at \(-2, 0,\) and \(3\) as follows:- For \(f(-2)\): Since \(-2\) is outside the domain \([-1, 2]\), \(f(-2)\) is undefined.- For \(f(0)\): \(0\) falls under the first condition, \(f(x) = 3x\). Thus, \(f(0) = 3 \times 0 = 0\).- For \(f(3)\): Since \(3\) is outside the domain \([-1, 2]\), \(f(3)\) is undefined.

Step 3: Graph the Function

To graph \(f\), we consider its piecewise nature:- For \(-1 \leq x < 1\), the graph is a line \(y = 3x\). This is a solid line starting from \(x = -1\) (where \(f(-1) = 3 \times (-1) = -3\)) to \(x = 1\) (where the value would be \(3\times1 = 3\), indicated with an open circle).- For \(1 \leq x \leq 2\), the graph is a line \(y = x + 1\), starting with a solid point at \(x=1\) (where \(f(1) = 2\)) to a solid point at \(x=2\) (where \(f(2) = 3\)).

Step 4: Determine Continuity

A function is continuous on its domain if there are no breaks or jumps. Here, at \(x=1\), the function has a jump: for \(x < 1\), \(f(1\)) would be expected to be \(3\), but \(f(1) = 2\) due to the second piece of the function. Hence, \(f\) is not continuous on its domain.

Key Concepts

Domain of a Function

The domain of a function encompasses all the possible input values, typically denoted as \( x \), that the function can accept. For piecewise functions, it's essential to examine each piece's condition. This means looking at which values of \( x \) are allowed within each segment. In our example function, \( f(x) \), there are two conditions:

• \(-1 \leq x < 1\)
• \(1 \leq x \leq 2\)

Step back and consider all values where the function is defined. We include the overlapping segments from both pieces to form our domain. The first segment covers up to, but not including \( x = 1 \), and the second segment includes \( x = 1 \) and extends to \( x = 2 \). Therefore, the domain of the piecewise function \( f \) is the interval \([-1, 2]\), inclusive at \( -1 \) and \( 2 \) but exclusive at \( 1 \) from the first piece.

Graphing Functions

Graphing functions, especially piecewise ones like \( f(x) \), involves plotting each piece of the function over their specified intervals. With piecewise functions, each section or part of the graph may follow a different rule or formula.

• For the first piece, where \(-1 \leq x < 1\), the equation is \( y = 3x \). Plot a line starting at \( x = -1 \) with \( f(-1) = -3 \) and end with an open circle at \( x = 1 \) to display that \( x = 1 \) is not included.
• The second piece, defined by \( y = x + 1 \) for \( 1 \leq x \leq 2 \), begins with a solid dot at \( x = 1 \) (where \( f(1) = 2 \)) and ends at \( x = 2 \) with a solid dot (where \( f(2) = 3 \)).

Paying careful attention to the endpoints—whether points are open or closed—ensures accuracy in reflecting which values of \( x \) the function includes.

Function Continuity

Function continuity refers to whether a graph is unbroken and smooth along its domain. For a function to be continuous, there should not be any jumps or breaks when transitioning from one piece to another. A function is continuous over an interval if, within that interval, there are no abrupt jumps or gaps between the points.In the case of \( f(x) \), the function is not continuous at \( x = 1 \). As you approach \( x = 1 \) from the left, \( f(x) \) per the first piece, would lead you to expect a value of \( 3 \). However, the function actually "jumps" to \( f(1) = 2 \) because of the second condition. This resulted in a break at this point, emphasizing that \( f(x) \) fails to meet the criteria for continuity across that interval.Identifying these discontinuities is essential as they often reveal critical insights about the behavior of function values at pinpointed segments.