Problem 22 - Calculus: Early Transcendentals

Question

Explain why the function is discontinuous at the given number \( a \). Sketch the graph of the function. \( f(x) = \left\\{ \begin{array}{ll} \dfrac{2x^2 - 5x - 3}{x - 3} & \mbox{if \) x \neq 3 \(} \hspace{30mm} a = 3\\\ 6 & \mbox{if \) x = 3 \(} \end{array} \right.\)

Step-by-Step Solution

Answer

The function is discontinuous at \(x = 3\) because \( \lim_{x \to 3} f(x) = 7 \neq f(3) = 6 \).

Step 1: Identify the Function Behavior

The function is defined as a piecewise function. For \(x eq 3\), it is given by \(f(x) = \dfrac{2x^2 - 5x - 3}{x - 3}\). At \(x = 3\), \(f(x) = 6\). To determine continuity at \(x = 3\), we need to check the limit of the function as \(x\) approaches 3 and the function value at \(x = 3\).

Step 2: Simplify the Algebraic Expression

Factor the quadratic expression in the numerator of \(f(x) = \dfrac{2x^2 - 5x - 3}{x - 3}\). The numerator \(2x^2 - 5x - 3\) can be factored as \((2x + 1)(x - 3)\), making the expression \(f(x) = \dfrac{(2x + 1)(x - 3)}{x - 3}\) when \(x eq 3\). Simplifying gives \(f(x) = 2x + 1\) for \(x eq 3\).

Step 3: Evaluate the Limit of the Function

Calculate the limit of \(f(x) = 2x + 1\) as \(x\) approaches 3: \[ \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7. \]

Step 4: Compare the Function Value and the Limit

At \(x = 3\), the function value is \(f(3) = 6\). Compare this to the limit as \(x\) approaches 3, which is 7. For continuity at a point, the function value must equal the limit; here, \(6 eq 7\).

Step 5: Conclusion of Discontinuity

Since the function value at \(x = 3\) (6) does not equal the limit as \(x\) approaches 3 (7), the function \(f(x)\) is discontinuous at \(x = 3\).

Step 6: Sketch the Graph

Sketch the graph of \(f(x)\). For \(x eq 3\), graph the line \(y = 2x + 1\) with a hole at \(x = 3\). Mark a distinct point at \((3, 6)\) since \(f(3) = 6\). The graph of the function has a jump from 7 (the limit value) to 6 (the defined function value at \(x = 3\)).

Key Concepts

Understanding Limits

Limits are a fundamental concept in calculus used to determine the behavior of a function as the input approaches a particular point. In the problem given, you need to calculate the limit of a function as it approaches a specific value, which was 3 in this case. The process involves simplifying the function expression wherever possible and then substituting the point into the simplified expression to find the limit. To illustrate, when simplifying the given piecewise function for values approaching 3, we made use of factoring the quadratic expression in the numerator and canceling the common factor with the denominator. This step brought the expression to a simple linear form, making it easier to evaluate the limit. Calculating the limit is crucial in understanding whether a function is continuous at a point. Remember, a function is continuous at a point when its limit as it approaches that point equals the function's exact value at that point.

Piecewise Functions Explained

Piecewise functions are defined by different expressions depending on the input values, capturing various behaviors within one function definition. They are particularly useful for modeling scenarios that have abrupt changes or different rules in different contexts. In the problem provided, the function is piecewise because it behaves differently for a specific value of the input, specifically when \( x = 3 \).For instance, our function assigns a different operation when \( x = 3 \). It provides a specific value (6), separate from the behavior of other \( x \) values which followed the linear equation derived from simplifying the original expression. Understanding how each piece of the function works is vital to find where discontinuities may exist, like the one seen at \( x = 3 \) in this problem.

Graph Sketching Basics

Graph sketching involves visually representing a function's behavior on a coordinate plane, helping to quickly identify patterns such as continuity or discontinuity. For the function in the exercise, graphing reveals the important aspects of its structure.The key points in sketching this function included graphing the linear part, \( y = 2x + 1 \), wherever \( x eq 3 \). It's important to leave a hole on the graph at \( x = 3 \) because the function does not continue in the same manner at that point. Furthermore, marking a unique point where \( x = 3 \) with the value of 6 showcases the specific definition of the piecewise function at that point. The graph will display a jump from the limit value 7 (as \( x \) approaches 3) to 6 at \( x = 3 \), indicating the discontinuity.

Working with Quadratic Expressions

Quadratic expressions are algebraic expressions that include a term with the variable squared. They appear frequently in functions, requiring familiarity with techniques such as factoring for simplification. In this exercise, we encountered a quadratic expression in the piecewise function: \( 2x^2 - 5x - 3 \). The goal was to factor it to assist in simplifying the function for calculating the limit. We managed to break it down into \((2x + 1)(x - 3)\), allowing us to cancel out the \( x - 3 \) from the denominator. This simplification reduced the expression to a linear form.Practicing factoring helps in recognizing patterns that allow for these simplifications, making it easier to handle higher-order algebraic expressions and evaluate functions effectively.