Question

Show that the following functions have a removable discontinuity at the given point. See Exercises \(95-96\). $$f(x)=\frac{x^{2}-7 x+10}{x-2} ; x=2$$

Short answer

Question: Show that the function $$f(x)=\frac{x^{2}-7x+10}{x-2}$$ has a removable discontinuity at x = 2. Answer: After simplifying the given function, the simplified function is continuous at x = 2, which indicates that the original function has a removable discontinuity at x = 2.

Step-by-step solution

Factor the numerator

We have the function $$f(x)=\frac{x^{2}-7 x+10}{x-2}$$. To factor the numerator, we look for two numbers that multiply to 10 and add to -7. These numbers are -5 and -2. So, the factored form of the numerator is: $$(x-5)(x-2)$$ Now, the function becomes: $$f(x)=\frac{(x-5)(x-2)}{x-2}$$

Simplify the function

Notice that both the numerator and denominator share a common factor of (x-2). We can cancel this out: $$f(x)=\frac{(x-5)(x-2)}{x-2}\Rightarrow f(x) = x-5$$

Check the continuity at x = 2

The simplified function is a linear function which is continuous for any value of x. Therefore, it is continuous at x = 2. Since the simplified function is continuous at x = 2, it proves that the original function had a removable discontinuity at this point.

Key concepts

Factorization

Understanding the concept of factorization is crucial when dealing with polynomial expressions, especially in identifying and resolving discontinuities in rational functions. Factorization involves breaking down a complex expression into simpler factors that, when multiplied together, yield the original expression. It's much like finding the building blocks that make up a structure.

In the exercise provided, the function’s numerator, \(x^2 - 7x + 10\), is a quadratic polynomial. The goal is to express this quadratic as a product of two binomials. To factorize a quadratic expression, we look for two numbers that multiply to the constant term (in this case, 10), and add up to the coefficient of the linear term (in this case, -7). Upon analyzing, we find that -5 and -2 satisfy these conditions, so the factored form becomes \((x-5)(x-2)\).

This process is crucial because it reveals common factors in the numerator and denominator of a rational function, which are key to identifying and addressing discontinuities. Without the ability to factor polynomials, it would be difficult to simplify rational expressions and fully understand their behavior.

Simplify Rational Expressions

To simplify rational expressions, we search for common factors in the numerator and denominator that can be canceled out. This is akin to reducing fractions in basic arithmetic; just as you would simplify \( \frac{8}{4} \) to 2 because 4 is a common factor of both the numerator and denominator, the same principle applies to algebraic expressions.

In the given exercise, after factorizing the numerator, our function had a common factor of \((x-2)\) in both the numerator and the denominator, allowing us to cancel it out and simplify the expression to just \(x-5\). However, it's important to note that canceling does not always mean the function behaves the same way as the simplified function at the point of cancellation. In this case, while the simplified expression is continuous, the original rational function is not defined at \(x=2\) due to division by zero. The simplification process leads us to uncover and rectify the removable discontinuity, clarifying the function’s behavior across its domain.

Continuity of Functions

The concept of continuity refers to a seamless, unbroken path when tracing a function on a graph. For a function to be continuous at a certain point, three conditions must be met: the function must be defined at that point, the limit of the function as it approaches the point must exist, and these two values must be equal.

When the exercise demonstrates a removable discontinuity at \(x=2\), it means that although there is a hiccup in the graph of the function at that point, this issue can be 'fixed' by altering the function slightly. Specifically, by removing the common factor causing the division by zero, the function becomes continuous across its entire domain. This is evident in the simplified version, where the linear function \(f(x) = x - 5\) is continuous everywhere, including at \(x=2\).

Therefore, the removable discontinuity is 'removed' when the simplification step is taken, bridging the gap that was present in the function's graph. This is why the term 'removable' is used - the discontinuity is not inherent to the function's nature but rather a result of how it's expressed.