MHB Finding Discontinuities: Multiply by Conjugate?

[RidiculousName]

I recently had to find what $$f(7)$$ equals if $$f(x) = \frac{x^2-11x+28}{x-7}$$. I first tried $$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$, and it seemed like a perfect fit since I eventually got to $$\frac{x^2(x-4)-49(x+4)}{x^2-49}=(x-4)(x+4)$$, but that gave me $$f(7)=33$$, instead of the right answer which was $$f(7)=3$$ since $$ \frac{x^2-11x+28}{x-7}=\frac{(x+7)(x-4)}{x-7}=x-4$$.

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 $$\frac{x+7}{x+7}$$ was the wrong approach here?

 

[Olinguito]

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 $\mathbb R\setminus\{7\}$.

 

[RidiculousName]

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 $x$.

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

 

[HOI]

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= 0and x- 7= 7- 7= 0. But the fact that x^2- 11x+ 28= 0 <b>tells</b> us that x- 7 <b>is</b> 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 <b>other</b> than 7, that is x- 4. The limit as x goes to 7 is 7- 4= 3.<br /> <br /> 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).

 

[Olinguito]

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.