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Chapter 12: Problem 132

For the following exercises, determine why the function \(f\) is discontinuous at a given point \(a\) on the graph. State which condition fails. $$ f(x)=\frac{x^{3}-27}{x^{2}-3 x}, \quad a=3 $$

Short Answer

Expert verified
The function is discontinuous at \(a=3\) because it is not defined at this point, creating a removable discontinuity.

Step by step solution

01

Simplify the Function Expression

We start with the given function: \[f(x) = \frac{x^3 - 27}{x^2 - 3x}\]We can simplify this expression. Notice that both the numerator and denominator can be factored. The numerator is the difference of cubes, and the denominator has a common factor. Factor them as follows:\[\text{Numerator: } x^3 - 27 = (x-3)(x^2 + 3x + 9)\]\[\text{Denominator: } x^2 - 3x = x(x-3)\]Now, substitute these factored forms back into the function:\[f(x) = \frac{(x-3)(x^2 + 3x + 9)}{x(x-3)}\]
02

Cancel the Common Factor

The simplified function now has a common factor of \((x-3)\) in both the numerator and the denominator. Cancel out this common factor:\[f(x) = \frac{x^2 + 3x + 9}{x}, \quad x eq 3\]The expression \((x-3)(..)\) cancels because it's present in both numerator and denominator. Note that \(x eq 3\) because division by zero is undefined and occurs when \(x = 3\).
03

Check for Discontinuity at a = 3

A function is discontinuous at a point for several reasons, such as if it is not defined, if it does not have a limit, or if the function's limit does not equal its value. Here, since cancelling caused a term \((x-3)\) to be removed:1. **Value of the function at \(a=3\):** Since division by zero makes it undefined at \(x=3\), \(f(3)\) is not defined.2. **Limit as x approaches 3:** Calculate \(\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 + 3x + 9}{x}\). - Direct substitution gives \(\frac{3^2 + 3 \cdot 3 + 9}{3} = \frac{9 + 9 + 9}{3} = 9\). - Therefore, \(\lim_{x \to 3} f(x) = 9.\)Since the function is not defined at \(x=3\) though it has a limit, based on the definition of continuity, it's not continuous at \(a=3\). This is a point of removable discontinuity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Cubes
The expression \(x^3 - 27\) in our problem is an example of a difference of cubes. This is a special form in algebra where we have the cube of one term subtracted by the cube of another. In this case, the formula for factoring such expressions is: \[a^3 - b^3 = (a-b)(a^2 + ab + b^2)\]
For our problem:
  • \(a^3 = x^3\), so \(a = x\)
  • \(b^3 = 27\), so \(b = 3\)
Applying the formula, \(x^3 - 27\) becomes \((x - 3)(x^2 + 3x + 9)\), breaking it down into easily manageable pieces.
This approach simplifies the function and helps to identify common factors.
Factoring
Factoring is breaking down an expression into a product of simpler factors. In the problem, we apply it in both the numerator and the denominator.
For the numerator, we used the difference of cubes method to factor \(x^3 - 27\) into \((x-3)(x^2 + 3x + 9)\).
In the denominator, we noticed a common factor: \(x^2 - 3x\). By factoring out an \(x\), we get \(x(x-3)\). This process of factoring allows us to simplify the expressions and find common elements.
  • Always look for common terms.
  • Simplifying helps make algebraic expressions easier to work with.
After factoring, we see the common factor \((x-3)\), which we can cancel from both the numerator and the denominator.
Limits
When we talk about limits in calculus, we're looking at what a function approaches as the input (or \(x\) value) gets closer to a certain number.
In our exercise, we were looking for the limit of \(f(x)\) as \(x\) approaches 3. After factoring and simplifying the function, removing the \(x-3\) factor lets us substitute \(x = 3\) directly into the simplified function.
  • The limit calculation showed that as \(x\) approaches 3, our function approaches 9.
  • Understanding limits helps determine continuity at a point.
Limits are essential to understand behavior at points where functions might normally be undefined. In this case, although the function was not defined exactly at \(x = 3\), it had a limit there.
Removable Discontinuity
A removable discontinuity occurs when a function is not defined at a point, but the limit exists as \(x\) approaches that point. In our exercise, this was observed at \(x = 3\).
Here are the key points:
  • The function \(f(x)\) is undefined at \(x=3\) because of division by zero.
  • Despite being undefined, the limit at \(x=3\) is 9.
This indicates a removable discontinuity because if we "remove" the discontinuity by redefining the function to include the limit at this point (make \(f(3) = 9\)), the function becomes continuous.
Recognizing removable discontinuities allows for the potential adjustment of a function so it's continuous over its entire domain.

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