Problem 72 - 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? $$ f(x)=\left\\{\begin{array}{ll} -2 & \text { if }-6 \leq x<-2 \\ 0 & \text { if }-2 \leq x<0 \\ 3 x & \text { if } \quad 0 \leq x \leq 4 \end{array}\right. $$

Step-by-Step Solution

Answer

The domain of \(f(x)\) is \([-6, 4]\). The values are \(f(-2) = 0\), \(f(0) = 0\), \(f(3) = 9\). It is not continuous due to jumps.

Step 1: Determine the Domain

Examine each piece of the piecewise function.- The first part is defined when \(-6 \leq x < -2\).- The second part is defined when \(-2 \leq x < 0\).- The third part is defined when \(0 \leq x \leq 4\).Unify these intervals to find the domain: \(-6 \leq x \leq 4\).

Step 2: Evaluate f(x) at Specific Points

Evaluate the function at given points based on the intervals:- For \(f(-2)\), the second part of the function applies, so \(f(-2) = 0\).- For \(f(0)\), the third part of the function applies, so \(f(0) = 3 \times 0 = 0\).- For \(f(3)\), the third part of the function applies, so \(f(3) = 3 \times 3 = 9\).

Step 3: Graph the Function

Draw the graph for each piece within its interval:- From \(-6\) to \(-2\), draw a horizontal line at \(y = -2\) and an open circle at \(x = -2\).- From \(-2\) to \(0\), draw a horizontal line at \(y = 0\) and open circles at both ends.- From \(0\) to \(4\), plot the line segment \(y = 3x\), with an open circle at \(x = 0\) and a closed circle at \(x = 4\).

Step 4: Determine Continuity

Check continuity across the pieces:- The function has a jump at \(-2\), moving from \(-2\) to \(0\).- The function is continuous between \(0\) and \(4\), but there is a jump at \(x=0\) from \(0\) to \(0\), meaning there is no change.Thus, the function is not continuous at \(-2\) but is continuous from \(0\) to \(4\).

Key Concepts

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable, usually represented as "x", that will result in a valid output when used in the function. For piecewise functions like the one in our exercise, the domain is determined by considering the specific intervals over which different expressions of the piecewise function are defined. For example, if a piece of the function is defined as \(-6 \leq x < -2\), only those values of \(x\) within this range are valid inputs for this part of the function.

To find the domain of the entire piecewise function, merge the valid intervals for each expression. In this exercise, we have three parts, each defined over a different range:

• The first part is defined from \(-6\) to \(-2\), excluding \(-2\).
• The second part is defined from \(-2\) to \(0\), again excluding \(0\).


• The third part is defined from \(0\) to \(4\), inclusive of both ends.

When these are combined, the unified domain of \(f(x)\) becomes \(-6 \leq x \leq 4\). This means \(x\) can take any value in this range to yield a valid output from the function.

Function Evaluation

Function evaluation is the process of determining the output of a function for specific input values. For piecewise functions, this involves selecting the appropriate piece of the function that corresponds to the input value. To evaluate our piecewise function at specific points, let's consider each interval of the piecewise function:

• For \(f(-2)\), we look at the interval \(-2 \leq x < 0\). Here, \(f(x) = 0\), so \(f(-2) = 0\).


• For \(f(0)\), we refer to the interval \(0 \leq x \leq 4\). Since \(f(x) = 3x\) in this interval, \(f(0) = 3 \times 0 = 0\).
• For \(f(3)\), again in the interval \(0 \leq x \leq 4\), we use \(f(x) = 3x\), so \(f(3) = 3 \times 3 = 9\).

By understanding which rule of the piecewise function applies to which interval, you can accurately evaluate the function at any specific point.

Graphing Functions

Graphing functions, especially piecewise ones, requires plotting each piece within its designated interval. This requires an understanding of how each piece behaves across its interval. Let's consider the pieces in our function:

• The first piece, \(y = -2\), is constant from \(-6\) to \(-2\), resulting in a horizontal line. At \(x = -2\), you would use an open circle to indicate that \(-2\) is not part of this interval.


• The second piece, \(y = 0\), spans from \(-2\) to \(0\). You sketch another horizontal line here with open circles at both \(-2\) and \(0\) because neither endpoint is included in this segment.
• The third piece, \(y = 3x\), represents a line that starts from the origin when \(x = 0\) to \(x = 4\). At \(x = 0\), you use an open circle and a filled circle at \(x = 4\) to show that this endpoint is included.

Connecting these sections carefully ensures that your graph accurately depicts the piecewise function.

Continuity of Functions

Continuity is an important property in functions, indicating that the graph of the function can be drawn without lifting the pencil off the paper. For a function to be continuous at a point, it must satisfy three conditions: the point must be part of the domain, the limit at that point must exist, and the limit must equal the function's value at that point.

In the case of piecewise functions like ours, let's inspect the continuity between its pieces:

• At \(x = -2\), there is a jump from \(-2\) to \(0\) as you switch from the first part to the second part, indicating discontinuity. This jump is visually represented by the shift from one horizontal line to another.
• Between \(0\) and \(4\), the function is a straightforward line segment (\(y = 3x\)); hence it is continuous over this range.

While the function has a discontinuity at \(x = -2\), it remains continuous in other sections, specifically where the line defined by \(y = 3x\) exists without breaks.