Problem 20 - Calculus Early Transcendentals

Question

\(15-20\) Explain why the function is discontinuous at the given number a. Sketch the graph of the function. \(f(x)=\left\\{\begin{array}{ll}{\frac{2 x^{2}-5 x-3}{x-3}} & {\text { if } x \neq 3} \\ {6} & {\text { if } x=3}\end{array}\right. \quad a=3\)

Step-by-Step Solution

Answer

The function is discontinuous at \( x = 3 \) because the limit at \( x = 3 \) is 7, but \( f(3) = 6 \).

Step 1: Understand the Problem

We need to determine if the function \( f(x) \) is discontinuous at \( x = 3 \). The function is defined in pieces: \( f(x) = \frac{2x^2 - 5x - 3}{x - 3} \) for \( x eq 3 \) and \( f(x) = 6 \) for \( x = 3 \).

Step 2: Simplify the Function for \( x \neq 3 \)

Factor the numerator of the expression \( \frac{2x^2 - 5x - 3}{x-3} \). The polynomial \( 2x^2 - 5x - 3 \) can be factored as \((2x+1)(x-3)\). So, for \( x eq 3 \), \( f(x) = \frac{(2x+1)(x-3)}{x-3} = 2x + 1 \).

Step 3: Evaluate the Limit as \( x \to 3 \)

Using the simplified function \( 2x + 1 \), compute the limit of \( f(x) \) as \( x \to 3 \):\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7. \]

Step 4: Compare the Limit and the Function Value at \( x = 3 \)

The computed limit as \( x \to 3 \), which is 7, does not match the defined function value \( f(3) = 6 \). This indicates that \( f(x) \) is discontinuous at \( x = 3 \).

Step 5: Sketch the Graph of the Function

Sketch two parts: 1. The line \( y = 2x + 1 \) with a hole at \( x = 3 \), since for \( x eq 3 \), \( f(x) = 2x + 1 \).2. A point at \( (3,6) \), indicating the piecewise defined value of the function at \( x = 3 \). The graph has a break at \( x = 3 \), indicating discontinuity.

Key Concepts

Piecewise Functions

Piecewise functions are defined by different expressions based on the input value of the variable. They are useful for modeling real-world scenarios where a different rule applies under different conditions. In the exercise provided, the function \( f(x) \) is defined as a piecewise function:

• \( f(x) = \frac{2x^2 - 5x - 3}{x - 3} \text{ for } x eq 3 \)
• \( f(x) = 6 \text{ for } x = 3 \)

This means that the function has one form when \( x \) is not equal to 3, and another form exactly at \( x = 3 \).
Piecewise functions often have different graphs for each piece of their definition, with breaks or jumps that can indicate discontinuities. Recognizing the structure of a piecewise function helps in analyzing these discontinuities, as well as in sketching its graph to visualize potential breaks.

Limits and Continuity

When we consider whether a function is continuous or not, we use the concepts of limits and continuity. A function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point.
In our exercise, we explored the function \( f(x) \). The limit as \( x \to 3 \) of the simplified function \( 2x + 1 \) is calculated as follows:

• \[ \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7 \]

However, the value of the function at \( x = 3 \) is given as 6. Because the limit as \( x \to 3 \) does not equal the function's value at 3, the function is discontinuous at \( x = 3 \).
Understanding the relationship between limits and continuity helps us determine where a function may "break" or "jump." This is crucial for identifying the precise points of discontinuity in piecewise functions.

Factorization of Polynomials

Factorization involves breaking down a complex polynomial into simpler terms that, when multiplied together, give back the polynomial. This is essential in simplifying functions to find limits, especially with piecewise functions.
For the piecewise function given in the exercise:

• The polynomial in the numerator \( 2x^2 - 5x - 3 \) can be factored as \((2x+1)(x-3)\).
• By factoring, the expression \( \frac{2x^2 - 5x - 3}{x-3} \) simplifies to \( 2x + 1 \) when \( x eq 3 \).

This simplification is possible because the factor \( (x-3) \) in the numerator and the denominator cancels out, leaving the linear expression \( 2x + 1 \).
Understanding factorization is vital as it not only helps simplify expressions but also plays a key role in determining the limits that reveal discontinuities in piecewise functions.