Problem 69 - 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 & \text { if }-5 \leq x \leq-1 \\ x+3 & \text { if }-1

Step-by-Step Solution

Answer

Domain: [-5, 5]; Evaluations: f(-2)=2, f(0)=3, f(3)=6; f is continuous on its domain.

Step 1: Determine the Domain

The domain of the function consists of all possible input values for which the function is defined. The function \( f(x) \) is defined in two pieces: the first piece is \( 2 \) when \( -5 \leq x \leq -1 \), and the second piece is \( x+3 \) when \( -1 < x \leq 5 \). Thus, the domain of \( f \) is from \(-5\) to \(5\), inclusive over \([-5,-1]\) and \(-1,5]\), or in interval notation, \([-5, 5]\).

Step 2: Evaluate f(-2), f(0), and f(3)

To evaluate \( f(-2) \), we use the expression for \( x \) in the interval \( -5 \leq x \leq -1 \), giving us \( f(-2) = 2 \).For \( f(0) \), the relevant interval is \( -1 < x \leq 5 \), so substitute \( 0 \) into \( x+3 \), yielding \( f(0) = 0 + 3 = 3 \).For \( f(3) \), use the interval \( -1 < x \leq 5 \), substituting \( 3 \) into \( x+3 \), giving \( f(3) = 3 + 3 = 6 \).

Step 3: Sketch a Graph of f

To graph \( f(x) \), notice that for \( -5 \leq x \leq -1 \), the graph is a constant horizontal line at \( y = 2 \). For \( -1 < x \leq 5 \), the graph is a line with a slope of \( 1 \) starting at point \((x=-1, y=2)\) and ending at \((x=5, y=8)\). Connect these with an open circle at \((x=-1, y=2)\) since it transitions to the second piece there, ensuring the graph is complete from \(x=-5\) to \(x=5\).

Step 4: Determine Continuity on Domain

A function is continuous if there are no breaks, jumps, or holes in its graph. \( f(x) \) is continuous on its entire domain because at \( x = -1 \), where the pieces meet, both limit from the left \( f(-1^-) = 2 \) and from the right \( -1 + 3 = 2 \) equal the value \( f(-1) = 2 \). There is no discontinuity or jump at \( x = -1 \).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial for determining the range of inputs that a function can accept. For a piecewise function like the one in our exercise, the domain is composed of the set of all possible input values that keep the function defined into all its separate rules.
For instance, our function has two distinct rules:

• For \(-5 \leq x \leq -1\), the function is constant and equals 2.
• For \(-1 < x \leq 5\), the function follows the expression \(x + 3\).

The domain is the union of these intervals, meaning it includes every number from \(-5\) to \(5\). In interval notation, it is expressed as \([-5, 5]\). Understanding the domain helps in knowing where the function can be graphically plotted or evaluated further.

Function Evaluation

Evaluating a function means finding the function's output for given input values. In our piecewise function, the approach varies based on which rule applies to a specific input value.
Let's see how this works in practice:

• For \(f(-2)\), since \(-5 \leq -2 \leq -1\), the function gives us \(2\).
• For \(f(0)\), \(-1 < 0 \leq 5\) applies, thus substitute into \(x + 3\), resulting in \(3\).
• Lastly, \(f(3)\) also uses the rule \(-1 < x \leq 5\), so \(3 + 3\) gives \(6\).

By understanding which piece of the function to use, function evaluation becomes straightforward and useful for predicting the behavior of the graph at specific points.

Graphing Functions

When graphing piecewise functions, you plot each part according to its definition on its domain section. Let's break down the graph of this function:

• For \(-5 \leq x \leq -1\), the graph is a horizontal line, since \(f(x) = 2\). Every value of \(x\) results in the same \(y\)-value, \(2\).
• For \(-1 < x \leq 5\), the graph is a line with a slope of 1. Start from \(x = -1, y = 2\), to the endpoint \((x = 5, y = 8)\).

Pay attention to the transition point at \(x = -1\). The line at \(-5 \leq x \leq -1\) finishes at \(x = -1\) with an open circle and simultaneously begins the next segment there with another open circle. This gives clarity that \(x = -1\) is repeated from the second piece.

Continuity of Functions

Determining continuity means ensuring the function has no interruptions like gaps, jumps, or holes within its domain. For piecewise functions, the critical area is at the interface between the rule changes.In this exercise, check \(x = -1\), where the two functions meet:

• Both the limit from the left at \(x = -1\) is \(2\) (from \(f(x) = 2\))
• And the limit from the right using \(-1 + 3\) is \(2\).

Finally, \(f(-1)\) is defined by continuation from \(f(x) = 2\).Since all values and limits agree, the function is continuous across its domain. This means the piecewise function is well-behaved, making it predictable and smooth throughout the given input range.