Problem 51 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(51 - 54 ,\) complete parts \(( a ) , (\) b) \(,\) and \(( c )\) for the piecewise-defined function. (a) Draw the graph of \(f .\) (b) Determine \(\lim _ { x \rightarrow c ^ { + } } f ( x )\) and \(\lim _ { x \rightarrow c ^ { - } } f ( x )\) (c) Writing to Learn Does \(\lim _ { x \rightarrow c } f ( x )\) exist? If so, what is it? If not, explain. $$c = 2 , f ( x ) = \left\\{ \begin{array} { l l } { 3 - x , } & { x < 2 } \\\ { \frac { x } { 2 } + 1 , } & { x > 2 } \end{array} \right.$$

Step-by-Step Solution

Answer

The graph of the piecewise function will be two separate lines intersecting at x = 2. The left limit of f(x) as x approaches from the left side (x < 2) in \(3-x\) is 1, while the right limit of f(x) as x approaches from the right side (x > 2) in \(\frac{x}{2} + 1\) is 2. Since both limits do not equal, the limit as \(x \rightarrow 2\) does not exist.

Step 1: Draw the Graph

Draw two separate graphs for each function, \(3 - x\) and \(\frac{x}{2} + 1\), respectively. But remember, \(3 - x\) is only valid for x < 2 and \(\frac{x}{2} + 1\) is valid for x > 2. Evaluate the functions at x=2 to ensure continuity.

Step 2: Determine the Right and Left Limits

Determine the left limit as \(x \rightarrow c \; (x \rightarrow 2^{-})\) and right limit as \(x \rightarrow c \; (x \rightarrow 2^{+})\). For the left limit, substitute \(x = 2\) in \(3-x\) to get 1 and for the right limit, substitute \(x = 2\) in \(\frac{x}{2} + 1\) to get 2.

Step 3: Determine the Existence of Limit

The overall limit as \(x \rightarrow c (x \rightarrow 2)\) exists only if the left limit equals the right limit. In this case, the limits are not equal (1 ≠ 2), therefore the limit as \(x \rightarrow 2\) does not exist.

Key Concepts

Graphing Piecewise Functions

Piecewise functions are special kinds of functions that have different expressions for different intervals in their domains. In other words, the rule for calculating the output value of the function changes depending on the input value you choose. To graph piecewise functions, you need to understand these different parts separately and plot them accordingly on the graph.

Consider the function given in the exercise:

• For values of \(x < 2\), the function is \(f(x) = 3 - x\).
• For values of \(x > 2\), the function is \(f(x) = \frac{x}{2} + 1\).

You would draw these two functions separately on a graph. Start by drawing the first part of the function \(3 - x\) for \(x < 2\). This is a straight line with a downward slope. Then, draw the second part \(\frac{x}{2} + 1\) for \(x > 2\), which is a straight line with an upward slope. Remember not to connect these two lines because the function is not defined at \(x = 2\). This creates a visual break or gap in the graph where \(x = 2\), which is called a discontinuity.

For graphing, ensure you use different styles, like a solid circle for included endpoints and open circle for excluded endpoints, to indicate whether a point is part of the function's graph.

Limits of Functions

In the context of piecewise functions, understanding limits helps us predict the behavior of a function as we approach a particular point from either side. For the piecewise function in our exercise, we are interested in finding the behavior of the function as we near \(x = 2\).

The left limit, \( \lim_{{x \to 2^-}} f(x) \), considers the approach from values of \(x\) less than 2. For \(x < 2\), the function \(f(x) = 3 - x\) yields 1 when \(x\) approaches 2. Thus, \( \lim_{{x \to 2^-}} f(x) = 1 \).

On the other hand, the right limit, \( \lim_{{x \to 2^+}} f(x) \), encompasses all values when approaching 2 from the right (where \(x > 2\)). Here, \(f(x) = \frac{x}{2} + 1\), so when approaching 2, the limit is \(2\). Therefore, \( \lim_{{x \to 2^+}} f(x) = 2 \).

Since these two limits are not equivalent, we conclude that \( \lim_{{x \to 2}} f(x) \) does not exist. This highlights a potential discontinuity at the breaking point.

Continuity of Functions

Continuity of a function at a particular point implies that the function is smooth and unbroken at that point. For a function to be continuous at a point, the limit as we approach that point from both sides must exist and must be equal to the function's value at that point. Furthermore, the function must be defined exactly at that specific point.

In our piecewise function:- To be continuous at \(x = 2\), three conditions are needed: 1. \(f(2)\) should be defined. 2. \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) \). 3. \( \lim_{{x \to 2}} f(x) = f(2) \).
In our scenario, neither the left-hand limit nor the right-hand limit are equal, and since \(f(x)\) is not defined at \(x = 2\) in our piecewise breakdown, there is a discontinuity at this point. Therefore, this function is not continuous at \(x = 2\). This discontinuity means that you would have to "lift your pencil" if you were trying to draw the graph, indicating a break or gap at the point.