# Finding Discontinuities: Multiply by Conjugate?

• MHB
• RidiculousName
In summary, the conversation discusses the process of finding the value of f(7) if f(x) is a rational function and the function's discontinuity at x=7. It is mentioned that the correct approach to calculating f(7) is to take the limit as x approaches 7, rather than substituting x=7 directly. The mistake of multiplying by the conjugate is also addressed and it is clarified that this is not necessary in finding the value of f(7). The importance of identifying factors and canceling them is emphasized.
RidiculousName
I recently had to find what $$\displaystyle f(7)$$ equals if $$\displaystyle f(x) = \frac{x^2-11x+28}{x-7}$$. I first tried $$\displaystyle \frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$, and it seemed like a perfect fit since I eventually got to $$\displaystyle \frac{x^2(x-4)-49(x+4)}{x^2-49}=(x-4)(x+4)$$, but that gave me $$\displaystyle f(7)=33$$, instead of the right answer which was $$\displaystyle f(7)=3$$ since $$\displaystyle \frac{x^2-11x+28}{x-7}=\frac{(x+7)(x-4)}{x-7}=x-4$$.

I don't have much experience with finding discontinuities, and I am confused because I had thought multiplying by by the conjugate was always the right way to go with these problems. Since I was wrong about that, I want to know how to tell what the wrong processes are with these types of problems so that I can do them correctly in the future. Is there a way I could've known that trying to multiply by $$\displaystyle \frac{x+7}{x+7}$$ was the wrong approach here?

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Hi RidiculousName.

If $f(x)=\dfrac{x^2-11x+28}{x-7}$, then $f(7)$ is not defined! Maybe you mean to find
$$\lim_{x\to7}f(x)$$
instead. Well, if $x\ne7$, then
$$f(x)\ =\ \frac{x^2-11x+28}{x-7}\ =\ \frac{(x-7)(x-4)}{x-7}\ =\ x-4$$
so $f(x)\to7-4=3$ as $x\to7$.

Note that the function defined by $g(x)=x-4$ is not the same function as $f(x)$. The former has domain $\mathbb R$ whereas $f(x)$ has domain $\mathbb R\setminus\{7\}$.

Olinguito said:
Hi RidiculousName.

If $f(x)=\dfrac{x^2-11x+28}{x-7}$, then $f(7)$ is not defined! Maybe you mean to find
$$\lim_{x\to7}f(x)$$
instead. Well, if $x\ne7$, then
$$f(x)\ =\ \frac{x^2-11x+28}{x-7}\ =\ \frac{(x-7)(x-4)}{x-7}\ =\ x-4$$
so $f(x)\to7-4=3$ as $x\to7$.

Note that the function defined by $g(x)=x-4$ is not the same function as $f(x)$. The former has domain $\mathbb R$ whereas $f(x)$ has domain $\mathbb R\setminus\{7\}$.

Thank you, I apologize for the mix-up about how f(7) isn't defined. I mentioned I have already discovered $f(x)\ =\ \frac{x^2-11x+28}{x-7}\ =\ \frac{(x-7)(x-4)}{x-7}\ =\ x-4$. I was asking if there was a way I could've known that my prior process of doing $\frac{x^2-11x+28}{x-7}$ wouldn't work.

Here are all of my steps.

$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$

$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$

$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$

$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$

$$=(x-4)(x+4)$$

plug in $7$ for $x$.

$$(7-4)(7+4)=7^2-16=33$$

With a rational function, f(x)= P(x)/Q(x), where P and Q are polynomials, at x= a there are 3 possibilities:
1) Q(a) is not 0 so that f(a)= P(a)/Q(a).

2) Q(a)= 0 but P(a) is not 0, neither f(a) nor lim as x goes to a of f(a) exists.

3) both P(a) and Q(a) are 0, f(a) does not exist but the limit might. That means that x- a must be factor of both P(a) and Q(a). Factor it out of numerator and denominator and cancel.

For example, to find the limit, as x goes to 7, of $$f(x)\frac{x^2- 11x+ 28}{x- 7}$$
I would first set x= 7 in the numerator and denominator: $$x^2- 11x+ 28= 49- 77+ 28= 0[tex] and [tex]x- 7= 7- 7= 0$$. But the fact that $$x^2- 11x+ 28= 0$$ tells us that x- 7 is a factor. Seeing the 28= 7(4), I would try $$(x- 7)(x- 4)= x^2- 7x- 4x+ 28= x^2- 7x+ 28$$ so that $$f(x)= \frac{x^2- 11x+ 28}{x- 7}= \frac{(x- 4)(x- 7)}{x- 7}$$ and, for all x other than 7, that is x- 4. The limit as x goes to 7 is 7- 4= 3.

There is no "multiplying by the conjugate" involved. In fact I am not sure what you mean by "conjugate" here. It is just a matter of recognizing that if P is a polynomial such that P(a)= 0 then x- a is a factor of P(x).

RidiculousName said:
Here are all of my steps.

$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$

$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$

$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$

$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$

$$=(x-4)(x+4)$$

plug in $7$ for $x$.

$$(7-4)(7+4)=7^2-16=33$$
You are doing a lot of unnecessary work for nothing at all. As Country Boy says, you do not have to multiply by anything extra. In any case, multiplying $f(x)$ by $\dfrac{x+7}{x+7}=1$ should give you $x-4$, not $(x-4)(x+4)$. You have made a mistake in your working – which in any case is immaterial because what you’re doing here is totally unnecessary.

PS:
$$\frac{x^3-4x^2-49x+196}{x^2-49}\ =\ \frac{x(x^2-49)-4(x^2-49)}{x^2-49}\ =\ \frac{(x^2-49)(x-4)}{x^2-49}\ =\ x-4$$
PROVIDED $x\ne\pm7$. By doing the unnecessary work, you’ve introduced an extra point of discontinuity, one which wasn’t there before.

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## 1. What is the purpose of multiplying by the conjugate when finding discontinuities?

When finding discontinuities, multiplying by the conjugate allows us to simplify complex expressions and identify any potential discontinuities. It is a useful technique in calculus and other mathematical applications.

## 2. How do I know when to use the conjugate when finding discontinuities?

The conjugate should be used when dealing with expressions that contain radicals or imaginary numbers. It is also helpful when dealing with fractions or expressions involving variables.

## 3. Can multiplying by the conjugate change the value of a function?

No, multiplying by the conjugate does not change the value of a function. It simply allows us to manipulate the expression and identify any potential discontinuities.

## 4. Is there a specific method for multiplying by the conjugate?

Yes, the method for multiplying by the conjugate is to multiply the numerator and denominator of the expression by the conjugate of the denominator. This will result in the elimination of any radicals or imaginary numbers in the denominator.

## 5. Can I use the conjugate to find all types of discontinuities?

No, the conjugate method is only effective for finding removable and jump discontinuities. Other types of discontinuities, such as infinite and oscillating, require different techniques for identification.

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