Problem 70 - 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? $$ f(x)=\left\\{\begin{array}{ll} 2 x+1 & \text { if }-3 \leq x<0 \\ x-1 & \text { if } \quad 0 \leq x \leq 3 \end{array}\right. $$

Step-by-Step Solution

Answer

(a) Domain: \([-3, 3]\). (b) \(f(-2) = -3\), \(f(0) = -1\), \(f(3) = 2\). (c) See graph. (d) Not continuous due to jump at \(x = 0\).

Step 1: Determine the Domain

The domain of a function is the set of all possible input values (x-values) that the function can accept. For this piecewise function, both pieces need to be considered:1. The first piece, \(2x+1\), is defined for \(-3 \leq x < 0\).2. The second piece, \(x-1\), is defined for \(0 \leq x \leq 3\).The domain of \(f(x)\) is therefore \(-3 \leq x \leq 3\).

Step 2: Evaluate f(-2)

To find \(f(-2)\), check which piece of the function it belongs to. Since \(-2\) is in the interval \(-3 \leq x < 0\), use \(f(x) = 2x + 1\):Substitute \(x = -2\):\[ f(-2) = 2(-2) + 1 = -4 + 1 = -3 \]

Step 3: Evaluate f(0)

To find \(f(0)\), check which piece of the function it belongs to. Since \(0\) is in the interval \(0 \leq x \leq 3\), use \(f(x) = x - 1\):Substitute \(x = 0\):\[ f(0) = 0 - 1 = -1 \]

Step 4: Evaluate f(3)

To find \(f(3)\), check which piece of the function it belongs to. Since \(3\) is in the interval \(0 \leq x \leq 3\), use \(f(x) = x - 1\):Substitute \(x = 3\):\[ f(3) = 3 - 1 = 2 \]

Step 5: Graph the Function

Draw a coordinate system with the x-axis ranging from -3 to 3.1. For the first piece \(y = 2x + 1\) over \(-3 \leq x < 0\): - At \(x = -3\), \(y = 2(-3) + 1 = -5\). - At \(x < 0\), approaching 0, \(y = 2(0) + 1 = 1\). - Plot the points \((-3, -5)\) to almost \((0, 1)\) with an open circle at \((0, 1)\).2. For the second piece \(y = x - 1\) over \(0 \leq x \leq 3\): - At \(x = 0\), \(y = 0 - 1 = -1\). - At \(x = 3\), \(y = 3 - 1 = 2\). - Plot the points \((0, -1)\) to \((3, 2)\).Connect these points with solid lines, showing a clear break at \(x = 0\).

Step 6: Analyze Continuity

A function is continuous if there are no breaks, jumps, or holes in the graph over its domain.Check continuity at \(x = 0\):- The left limit as \(x\) approaches 0 is \(1\) (from \(2x+1\)).- The right limit and function value at 0 is \(-1\) (from \(x-1\)).Since there is a jump discontinuity at \(x = 0\), the function is not continuous over its entire domain.

Key Concepts

Domain of a Function

In mathematics, the domain of a function is a critical concept that defines where the function is applicable. It represents all the possible input values (x-values) that one can plug into the function to get a real output. For piecewise functions, like the one in our exercise, the domain is determined separately for each piece or segment of the function.

For this specific function, we have two pieces with different conditions:

• The expression \(2x + 1\) is valid for \(-3 \leq x < 0\).
• The expression \(x - 1\) is valid for \(0 \leq x \leq 3\).

By combining these intervals, the domain of our piecewise function is all the x-values from \(-3\) to \(3\), inclusive at both ends. Understanding domains helps us figure out where a function "lives" on the x-axis, guiding us in function evaluation and graphing.

Function Evaluation

Function evaluation is the process of finding output values, or y-values, given specific input values of x. When dealing with piecewise functions, identifying the correct piece of the function to use is crucial. Each segment of a piecewise function may have its own formula, applicable only within certain ranges.

In our exercise, we evaluated the function at specific points:

• For \(x = -2\), since it's in the interval \(-3 \leq x < 0\), we use the formula \(2x + 1\). Calculating,the result is \(f(-2) = -3\).
• For \(x = 0\), it falls in the interval \(0 \leq x \leq 3\), so we use \(x - 1\). Hence, \(f(0) = -1\).
• For \(x = 3\), still within \(0 \leq x \leq 3\), use \(x - 1\), giving \(f(3) = 2\).

Correctly evaluating functions helps ensure accurate graphing and analysis of how a function behaves at specific points.

Graphing Functions

Graphing functions gives a visual understanding of how a function behaves across its domain. When graphing piecewise functions, each part is graphed separately based on the specified conditions and ranges.

To graph the given function, follow these steps:

• First, graph \(y = 2x + 1\) for \(-3 \leq x < 0\). This includes points from \((-3, -5)\) to almost \((0, 1)\), using an open circle at \((0, 1)\) since \(0\) is not included.
• Next, graph \(y = x - 1\) for \(0 \leq x \leq 3\), connecting points \((0, -1)\) to \((3, 2)\) with solid lines.

Drawing these segments accurately shows where the function changes its form, revealing any breaks or jumps and providing a clear picture of the function's overall behavior.

Continuity of Functions

Continuity in functions is about the smoothness of the graph; a continuous function does not have breaks, jumps, or holes within its domain. For a function to be considered continuous, the limit of the function from the left must equal the limit from the right at every point, and these must equal the function value at that point.

In our example, the function is analyzed for continuity at \(x = 0\). Here's what we observe:

• The left limit as \(x\) approaches zero (from \(-\)) using \(2x + 1\) is \(1\).
• The right limit as \(x\) approaches zero (from \(+\)) and the function value using \(x - 1\) is \(-1\).

Since \(f(0)\) has different values coming from left and right, there is a jump discontinuity at \(x = 0\). Therefore, the function is not continuous on its domain. This aspect of functions is crucial for determining where graphs are unbroken and helps in different applications where smoothness is required.