Problem 64 - Calculus and its applications

Question

Is the function given by \(f(x)=\frac{1}{x^{2}-6 x+8}\) continuous at \(x=3 ?\) Why or why not?

Step-by-Step Solution

Answer

The function is continuous at \( x = 3 \) because \( \lim_{{x \to 3}} f(x) = f(3) \).

Step 1: Identify the Function and Point of Interest

The function given is \( f(x) = \frac{1}{x^2 - 6x + 8} \). We need to check if this function is continuous at \( x = 3 \). Continuity at a point means the function is defined at that point, the limit exists at that point, and the limit equals the function's value.

Step 2: Factor the Denominator

The expression in the denominator is \( x^2 - 6x + 8 \). We will factor it to understand its zeros: \( x^2 - 6x + 8 = (x - 2)(x - 4) \). This tells us the function has potential discontinuities at \( x = 2 \) and \( x = 4 \) since the denominator becomes zero there.

Step 3: Check Function Definability at x = 3

Check if \( f(3) \) is defined by substituting \( x = 3 \) into the function:\[ f(3) = \frac{1}{3^2 - 6 \times 3 + 8} = \frac{1}{9 - 18 + 8} = \frac{1}{-1} = -1 \].The function is defined at \( x = 3 \).

Step 4: Determine Limit as x Approaches 3

We need to find \( \lim_{{x \to 3}} f(x) \) to determine continuity:Since \( x=3 \) is not one of the problematic points (where the denominator is zero), we can directly substitute \( x = 3 \) into the function:\[ \lim_{{x \to 3}} f(x) = \frac{1}{3^2 - 6 \times 3 + 8} = \frac{1}{-1} = -1 \].

Step 5: Compare Limit and Function Value

For \( f(x) \) to be continuous at \( x = 3 \), the function value and the limit at \( x = 3 \) need to be equal:\( \lim_{{x \to 3}} f(x) = f(3) = -1 \)Since the limit equals the function value, \( f(x) \) is continuous at \( x = 3 \).

Key Concepts

Understanding Limits

Limits help us describe the behavior of functions as they approach a certain point. To check a limit, we observe the value that a function gets closer to, as the variable approaches a specific point. In our exercise, we evaluated the limit of the function as its input, \(x\), nears 3.

• A limit can exist even if the function is not defined at a certain point.
• If the limit as \(x\) approaches a certain value equals the function's value at that point, the function can be continuous there.
• Using limits, we can ascertain the function’s behavior near potentially troublesome points like where the denominator is zero.

In the provided exercise, the limit of \(f(x) = \frac{1}{x^2 - 6x + 8}\) as \(x\) approaches 3 is evaluated. It directly substitutes \(x = 3\) into the function, showing the limit exists and matches the function's value.

The Role of Factoring Polynomials

Factoring polynomials is a critical step in simplifying expressions and finding denominators that could potentially lead to undefined functions. In this exercise, we factored the denominator \(x^2 - 6x + 8\) into \((x-2)(x-4)\).

• Factoring helps us identify zeros of the polynomial, where the expression becomes zero.
• These zeros indicate potential discontinuities because division by zero is undefined.
• By factoring, we uncovered that the expression has zeros at \(x = 2\) and \(x = 4\).

Thus, understanding how to factor helps us foresee where a function might not behave as expected, even if the exercise is at another point entirely, such as \(x = 3\) here.

Evaluating Functions at Specific Points

Evaluating a function at a specific point means substituting the value into the function to find its output. In the exercise, we evaluated the function \(f(x) = \frac{1}{x^2 - 6x + 8}\) at \(x = 3\). This step checks whether the function is defined there.

• If a function outputs a real number, it is defined at that point.
• A point where a function isn't defined can lead to discontinuity.
• For the given function, \(f(3) = -1\), ensuring the function is defined and can be evaluated at \(x = 3\).

Evaluating functions at specific points is essential in determining continuity because it reveals whether the function has a valid output at that point.

Identifying Discontinuities in Functions

Discontinuities occur when a function cannot be smoothly drawn without lifting the pen. They are critical in understanding why certain points cannot be included in the domain of a function.

• Discontinuity at a point means the function is not continuous there.
• Discontinuities often arise when the denominator of a fraction is zero, leading to an undefined function.
• In this exercise, \(x = 2\) and \(x = 4\) are potential discontinuities because the denominator zeroes out there.

However, \(x = 3\) does not cause a division by zero, hence no discontinuity here. Understanding discontinuities can clarify the limits and where a function does not behave in typical fashion.