Question

Show that the following functions have a removable discontinuity at the given point. See Exercises \(91-92\) $$f(x)=\frac{x^{2}-7 x+10}{x-2} ; x=2$$

Short answer

Question: Show that the function $$f(x) = \frac{x^2 - 7x + 10}{x - 2}$$ has a removable discontinuity at $$x = 2$$. Answer: The function $$f(x) = \frac{x^2 - 7x + 10}{x - 2}$$ has a removable discontinuity at $$x = 2$$ since its limit as x approaches 2 is -3, and there exists an equivalent function that is continuous at that point, $$g(x) = \begin{cases} x-5, & \quad \text{if } x \neq 2\\ -3, & \quad \text{if } x = 2 \\\end{cases}$$.

Step-by-step solution

Factor the numerator

Factor the quadratic expression $$x^2 - 7x + 10$$ to obtain $$(x - 2)(x - 5)$$. So, the function can be written as: $$f(x) = \frac{(x - 2)(x - 5)}{x - 2}$$

Simplify the function

As long as $$x\neq2$$, we can cancel the common factor $$(x - 2)$$ in the numerator and the denominator: $$f(x) = x - 5$$ , when $$x\neq2$$.

Find the limit of the function as x approaches 2

Now that we have simplified the function, we can find the limit as x approaches 2: $$\lim_{x\to2}f(x)=\lim_{x\to2}(x-5)$$ Substitute $$x=2$$ to find the limit: $$\lim_{x\to2}(x-5) = 2 - 5 = -3$$ The limit of the function as x approaches 2 is -3.

Define equivalent function with removable discontinuity

Now that we know the limit of the function exists at $$x=2$$, we can define an equivalent function that is continuous at that point: $$g(x) = \begin{cases} x-5, & \quad \text{if } x \neq 2\\ -3, & \quad \text{if } x = 2 \\ \end{cases} $$ This function is continuous everywhere, including at $$x=2$$. In conclusion, the function $$f(x) = \frac{x^2 - 7x + 10}{x - 2}$$ has a removable discontinuity at $$x = 2$$ since its limit exists at that point, and there exists an equivalent function that is continuous at that point.

Key concepts

Rational Functions

Rational functions are mathematical expressions that involve two polynomials in the form of a fraction. In essence, they look something like \( \frac{P(x)}{Q(x)} \), where both the numerator \( P(x) \) and the denominator \( Q(x) \) are polynomials. These functions can present complex behavior: they may have discontinuities, usually occurring when the denominator is zero.
In the provided exercise, we have a rational function \( f(x) = \frac{x^2 - 7x + 10}{x - 2} \). The most crucial aspect to note is where this function is undefined—specifically when \( x = 2 \), as the denominator becomes zero, potentially leading to a division by zero error, which is undefined in mathematics.
The significance of identifying such points is to determine whether these discontinuities are removable, which means we can rewrite or simplify the function into a form where the discontinuity no longer exists.

Limits

A limit helps us understand the behavior of a function as it approaches a specific point, even if the function might not be properly defined at that point itself. In simpler terms, it tells us what "y-value" a function is getting closer to as \( x \) gets closer and closer to a particular value.
In our exercise, the limit of \( f(x) \) as \( x \) approaches 2 is crucial for establishing whether the discontinuity is removable. The step-by-step solution simplifies the function \( f(x) \) by canceling the common factor \((x - 2)\) in both the numerator and denominator. This yields \( f(x) = x - 5 \) as long as \( x eq 2 \).
We then calculate \( \lim_{x \to 2}(x - 5) \). By directly substituting \( x = 2 \), the limit of the function equals \(-3\). Because this limit exists, we have a solid basis for asserting that the discontinuity at \( x = 2 \) is removable.

Factoring Quadratics

Factoring is a crucial algebraic skill, especially when working with quadratics. It allows us to express a quadratic in its simplest form, potentially simplifying complex expressions. For the function \( f(x) = \frac{x^2 - 7x + 10}{x - 2} \), this step is essential to identifying removable discontinuities.
To factor the numerator \( x^2 - 7x + 10 \), we look for two numbers that multiply to 10 (the constant term) and add up to -7 (the coefficient of the middle term). These numbers are -2 and -5. Thus, we factor the quadratic as \((x - 2)(x - 5)\).
Factoring, in this case, reveals the common factor \( (x - 2) \) present in both the numerator and denominator. Simplifying the function by cancelling out this common factor allows us to handle the point where the function is otherwise undefined, effectively "removing" the discontinuity. This process showcases the power of factoring in analyzing and simplifying rational functions.