Problem 70 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The function that gives the cost of a certain commodity is defined by $$ C(x)=\left\\{\begin{array}{ll} 5 x & \text { if } 0

Step-by-Step Solution

Answer

The function C(x) is discontinuous at \(x = 10, 30, \text{ and } 60 \). To sketch the graph: 1. For \(0 < x < 10\), plot \(C(x) = 5x\) as a line segment starting with an open circle at (0,0) and ending with an open circle at (10,50). 2. For \(10 \leq x < 30\), plot \(C(x) = 4x\) as a line segment starting with a closed circle at (10,40) and ending with an open circle at (30,120). 3. For \(30 \leq x < 60\), plot \(C(x) = 3.5x\) as a line segment starting with a closed circle at (30,105) and ending with an open circle at (60,210). 4. For \(x \geq 60\), plot \(C(x) = 3.25x\) as a line segment starting with a closed circle at (60,195) and extending in the direction of increasing x.

Step 1: Identify the equations for each range

First, let's make note of the equations used in each interval: 1. For \( 0 < x < 10 \), the equation is \( C(x) = 5x \). 2. For \( 10 \leq x < 30 \), the equation is \( C(x) = 4x \). 3. For \( 30 \leq x < 60 \), the equation is \( C(x) = 3.5x \). 4. For \( x \geq 60 \), the equation is \( C(x) = 3.25x \).

Step 2: Identify the end points

For each interval, let's identify the end points: 1. \( x = 10 \) is the endpoint of the first function (\( 5x \)) and the second function (\( 4x \)). 2. \( x = 30 \) is the endpoint of the second function (\( 4x \)) and the third function (\( 3.5x \)). 3. \( x = 60 \) is the endpoint of the third function (\( 3.5x \)) and the fourth function (\( 3.25x \)).

Step 3: Evaluate the functions at the end points

Next, evaluate each function at their end points: 1. At \( x = 10 \), the first function gives \( C(x) = 5(10) = 50 \), and the second function gives \( C(x) = 4(10) = 40 \). 2. At \( x = 30 \), the second function gives \( C(x) = 4(30) = 120 \), and the third function gives \( C(x) = 3.5(30) = 105 \). 3. At \( x = 60 \), the third function gives \( C(x) = 3.5(60) = 210 \), and the fourth function gives \( C(x) = 3.25(60) = 195 \).

Step 4: Identify the points of discontinuity

A function is discontinuous at a point where it is not continuous. Since the function changes its equation at each endpoint (10, 30, 60), it implies that the function is discontinuous at these points.

Step 5: Sketch the graph of the function

To sketch the graph: 1. For \( 0 < x < 10 \), plot the equation \( C(x) = 5x \) as a line segment that starts with an open circle at (0, 0) and ends with an open circle at (10, 50). 2. For \( 10 \leq x < 30 \), plot the equation \( C(x) = 4x \) as a line segment that starts with a closed circle at (10, 40) and ends with an open circle at (30, 120). 3. For \( 30 \leq x < 60 \), plot the equation \( C(x) = 3.5x \) as a line segment that starts with a closed circle at (30, 105) and ends with an open circle at (60, 210). 4. For \( x \geq 60 \), plot the equation \( C(x) = 3.25x \) as a line segment that starts with a closed circle at (60, 195) and extends in the direction of increasing x. The graph of the function should show discontinuities at \( x = 10, 30 \text{, and } 60 \), where there are open circles on one side of the interval and a closed circle on the other side.

Key Concepts

Discontinuity

In mathematics, a function is said to be discontinuous at a certain point if there is an abrupt change in its value. For a piecewise function like the one described in the exercise, discontinuity often occurs at the boundaries between different intervals. This particular function changes its equation at three key points: \(x = 10\), \(x = 30\), and \(x = 60\). At these points, the function fails to be continuous because the left-hand limit does not equal the right-hand limit.

Consider what happens as \(x\) approaches 10 from the left, where the value is calculated using \(C(x) = 5x\). Here, \(C(10^-) = 50\). However, approaching from the right uses \(C(x) = 4x\), giving \(C(10^+) = 40\). This jump suggests a discontinuity.

The same kind of disruption occurs at 30 and 60. For instance, when \(x\) nears 30, \(C(30^-) = 120\) versus \(C(30^+) = 105\). By understanding discontinuities, students can anticipate where a piecewise function will have breaks or jumps on its graph and function evaluations can vary based on direction of approach.

Graph Sketching

Graph sketching for piecewise functions involves careful plotting of each interval defined by the function's equations. For the given exercise, each interval corresponds to a straight line segment with different slopes; these slopes are determined by the linear functions defined over each specific range.

• For \(0 < x < 10\), plot \(C(x) = 5x\) starting from an open circle at (0, 0) to an open circle at (10, 50), indicating the endpoints are not included in this segment.
  
• For \(10 \leq x < 30\), graph \(C(x) = 4x\) from a closed circle at (10, 40) to an open circle at (30, 120). This shows inclusion at \(x = 10\) but not at \(x = 30\).
  
• For \(30 \leq x < 60\), draw the line representing \(C(x) = 3.5x\) with a closed start at (30, 105) to an open end at (60, 210).
  
• Finally, for \(x \geq 60\), plot \(C(x) = 3.25x\) extending from (60, 195) onwards.

Connecting these segments carefully, taking note of open versus closed circles, helps visualize the function's behavior and its points of discontinuity.

Function Evaluation

Evaluating a piecewise function involves identifying the correct equation for the interval that includes the point of interest. For example, to find \(C(x)\) specifically at \(x = 25\), one must select the appropriate sub-function based on the range that \(x\) falls into. Since 25 fits in \(10 \leq x < 30\), we use \(C(x) = 4x\), resulting in \(C(25) = 4 \cdot 25 = 100\).

When facing endpoints like \(x = 30\), we examine the interval definitions: \(x = 30\) is included in the \(30 \leq x < 60\) range, not the \(10 \leq x < 30\) one. We thus evaluate using \(C(x) = 3.5x\) to get \(C(30) = 105\).

• To evaluate a piecewise function at any point:
  
  • Determine which min-max conditions \(x\) satisfies;
    
  • Select the correct equation for this interval;
    
  • Substitute \(x\) in this equation.
    
• Be mindful of open and closed intervals, as this will affect which values are included.
  

Consistency in evaluating function components ensures the correct application of piecewise functions across varying intervals.