Question

Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{cl}\frac{x^{2}-64}{x^{2}-11 x+24} & x \neq 8 \\ 5 & x=8\end{array}\right.\) (a) \(x=0\) (b) \(x=8\)

Short answer

The function is continuous at \( x = 0 \), but not at \( x = 8 \).

Step-by-step solution

Understand the condition of continuity

A function \( f \) is continuous at a point \( x = a \) if the following three conditions are satisfied: 1) \( f(a) \) is defined, 2) \( \lim_{{x \to a}} f(x) \) exists, and 3) \( \lim_{{x \to a}} f(x) = f(a) \). We will check these conditions at each point.

Check continuity at \( x = 0 \)

Substitute \( x = 0 \) in the function. The function is not explicitly defined for \( x = 0 \), but it takes the form of the rational function. Substitute \( x eq 8 \) form: \[ f(0) = \frac{0^2 - 64}{0^2 - 11 \times 0 + 24} = \frac{-64}{24} = -\frac{8}{3} \].Next, find the limit: \[ \lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{x^2 - 64}{x^2 - 11x + 24} = -\frac{8}{3} \].Since \( f(0) = -\frac{8}{3} \) and the limit exists and matches \( f(0) \), the function is continuous at \( x = 0 \).

Evaluate and simplify function for limits at \( x = 8 \)

The function provided does not directly give \( f(x) \) for \( x eq 8 \) to exist directly. Simplify the rational expression to find limits at \( x = 8 \):\[ f(x) = \frac{x^2 - 64}{x^2 - 11x + 24} = \frac{(x-8)(x+8)}{(x-3)(x-8)} \].For \( x eq 8 \), this simplifies to \( f(x) = \frac{x+8}{x-3} \).

Check continuity at \( x = 8 \)

Calculate \( \lim_{{x \to 8}} f(x) \) using the simplified expression:\[ \lim_{{x \to 8}} \frac{x+8}{x-3} = \frac{8+8}{8-3} = \frac{16}{5} \].Now compare with \( f(8) \) given as \( 5 \). Since \( \lim_{{x \to 8}} f(x) = \frac{16}{5} eq 5 \), the function is not continuous at \( x = 8 \) because the limit does not equal the function value.

Key concepts

Understanding Limits in Continuity

When we speak about the continuity of a function, limits play an essential role. A function is continuous at a point if it does not "break" or "jump" as you approach this point, and an important part of this definition involves limits.

Limits help us understand what the value of a function is as it approaches a particular point. Mathematically, if we denote a function by \( f(x) \), then the limit of \( f(x) \) as \( x \) approaches \( a \) is expressed as \( \lim_{{x \to a}} f(x) \). For continuity at a point \( a \), not only should this limit exist, but it must also equal the actual function value at \( f(a) \).

Here's how it works:

• If \( \lim_{{x \to a}} f(x) \) exists, that means the function approaches some specific value as \( x \) gets close to \( a \).


• If \( f(a) \) is defined, this ensures there's a definitive value at \( a \).


• Finally, if \( \lim_{{x \to a}} f(x) = f(a) \), it means the function doesn't jump or have any gaps at \( a \).

If any of these conditions are not met, the function is not continuous at the point.

Piecewise Functions and Their Continuity

Piecewise functions are special in that they can behave differently in different regions of their domain. These functions are defined by different expressions based on different intervals of the input variable. However, for a piecewise function to be continuous, each piece must "connect" smoothly at the boundaries.

Consider a function like \( f(x) \) shown as a piecewise function:

• \( f(x) = \frac{x^2 - 64}{x^2 - 11x + 24} \) for \( x eq 8 \)


• \( f(x) = 5 \) for \( x = 8 \)

When assessing such functions for continuity, we must check the transition points where the definition of the function changes.

For example, at \( x=8 \), the two conditions must satisfy the definition of continuity:

• The function's limit as \( x \to a \) from both sides should align with \( f(a) \).


• If it doesn't, the function isn't continuous at that juncture.

This ensures a seamless, connected flow without sudden changes.

Rational Functions and Their Simplification

Rational functions are defined as the ratio of two polynomials. A common feature is points where they become undefined, often leading to breaks in continuity. Simplifying these functions is crucial to understanding their behavior and finding limits. Simplification often involves factoring to cancel out common terms that may cause these functions to be undefined at certain points.

Take \( f(x) = \frac{x^2 - 64}{x^2 - 11x + 24} \), a rational function. By factoring both the numerator and the denominator, we get:

\[ f(x) = \frac{(x-8)(x+8)}{(x-3)(x-8)} \]

Upon canceling the \( (x-8) \) term, which causes the initial undefined state when \( x = 8 \), we simplify it to \( f(x) = \frac{x+8}{x-3} \) for \( x eq 8 \).

This simplification shows the true behavior of the function around \( x = 8 \) and assists in calculating limits smoothly. Factoring and cancellation are key strategies to remove these indeterminate forms and understand where limits exist and where they might cause discontinuity.