# Possible values an expression can take : ##\dfrac{x^2-x-6}{x-3}##

• brotherbobby
In summary, the conversation discusses the process of finding the domain and range of a rational function, specifically the expression ##\frac{x^2-x-6}{x-3}##. The issue of finding the domain of an "inverse" function is addressed and it is noted that the inverse is not a true function. The use of the discriminant in finding the domain is also discussed, with the reminder to be aware of potential discontinuities in the original function.
brotherbobby
Homework Statement
Find the possible values that the following expression can take : ##\dfrac{x^2-x-6}{x-3}##
Relevant Equations
1. For a (rational) function of the form ##y=\frac{f(x)}{g(x)}##, we should have the function ##g(x)\ne 0\; \forall \; x##. This would determine the domain of the function ##\mathscr{D}##
2. If we can invert the function ##y=f(x)## to obtain ##x=f^{-1}(y)##, the allowable values of ##y## would determine the range of the function ##\mathscr{R}##.
Problem statement :
Let me copy and paste the problem as it appears in the text :
Attempt 1 (from text) : The book and me independently could solve this problem. I copy and paste the solution from the book below.

Attempt 2 (my own) : The problem should afford a solution using the second idea I put in the Relevant Equations above - namely that the range of the function can be obtained by inverting it. So let ##y=x^2−x−6x−3⇒x^2−x−6=xy−3y⇒x^2−x(1+y)+3(y−2)=0##. For the values of x to be real ##(x∈R)##, the discriminant of the function ##D=b^2−4ac≥0⇒(1+y)^2−12(y−2)≥0⇒y^2−10y+25≥0⇒(y−5)^2≥0##. But the square of a number is always greater than 0. Hence y, which is the expression, can take all values : ##y∈R##.

Issue : Clearly my answer misses out on y≠5. This could have been achieved if the discriminant D>0. However, the discriminant can be equal to 0 also, in which case we would only have one real root were the quadratic expression was set to zero.

A help or a hint would be welcome.

Delta2
brotherbobby said:
But the square of a number is always greater than 0.
Not quite. The square of a number is always greater than or equal to 0.
brotherbobby said:
the range of the function can be obtained by inverting it. So let ##y=x^2−x−6x−3⇒x^2−x−6=xy−3y⇒x^2−x(1+y)+3(y−2)=0##.
Your first equation is incorrect, since you omitted the LaTeX to make it a fraction. It took me awhile to figure out how you got the 2nd equation from the first. First equation should be:
##y = \frac{x^2 - x - 6}{x - 3}##

To get to the 2nd equation, you need to multiply both sides by x -3, which presumes that ##x \ne 3##. If y = 5, x will equal 3.

Last edited:
Delta2
Notice the function is not defined for ##x=3 ##, as you noted. You can define it in any way you wish. There is one choice for ##x## that would make it continuous.

I want to respond to @Mark44 with apologies for my typing in ##\LaTeX##. Let me restate the problem and present both solutions as I attempted in my post # 1 above.

Problem statement : Find all the possible values that the expression can take : ##\dfrac{x^2-x-6}{x-3}##.

First Attempt : I paste my written solution below which agrees with the text. (I hope it's readable)

Answer : The expression can take values ##\boxed{\mathbb{R} - \{5\}}##.

Second attempt : This is the problematic part. By reason, if I substitute ##y=\dfrac{x^2-x-6}{x-3}## and then express ##x=f^{-1}(y)##, then the allowable values of ##y## is my domain.

I copy and paste, hoping it's readable.

The solution I now get is that ##y\in R##. Of course I am aware that ##x\ne 3##. However, that condition, or equivalently ##y \ne 5## must come automatically from the calculation above, but it doesn't.

Please tell me where am going wrong.

brotherbobby said:
The solution I now get is that ##y\in R##. Of course I am aware that ##x\ne 3##. However, that condition, or equivalently ##y \ne 5## must come automatically from the calculation above, but it doesn't.

Please tell me where am going wrong.
In your work toward solving for x in terms of y, you focused exclusively on the discriminant. The full equation involving x and y is ##x = \frac{1 + y \pm \sqrt{(y - 5)^2}}{2}##. Of course the discriminant is nonnegative for all real y, but if y = 5, x = 3, which is not in the domain of the original function.
So, as before, the domain for the "inverse" (which is not a function) is ##\{y \in \mathbb R, y \ne 5 \}##

I agree with you, but this means that it is not a straightforward thing to ascertain the domain of a function ##f(x)## by simply putting it equal to some variable ##y (=f(x))## and then inverting the function to see what appropriate values ##y## can then take. One might have to, at times, also look for the domain for the function ##(\text{values of x})## and then compute the values in the range (##y##) for those forbidden domain values of ##x##.

My example above in this thread is a good one. There is nothing in the inverted function of ## = f^{-1}(y)## to suggest that the value ##y=5## is forbidden. And yet it is, separately seen from the function ##f(x)## itself.

Thank you for your interest.

A case of indiscriminate use of the discriminant!

brotherbobby said:
I agree with you, but this means that it is not a straightforward thing to ascertain the domain of a function ##f(x)## by simply putting it equal to some variable ##y (=f(x))## and then inverting the function to see what appropriate values ##y## can then take. One might have to, at times, also look for the domain for the function ##(\text{values of x})## and then compute the values in the range (##y##) for those forbidden domain values of ##x##.
Sure, it's not 100% straightforward, but there are things you can watch out for. In this case, you multiplied both sides of the equation ##y = \frac{x^2 - x - 6}{x - 3}## by x - 3 to get to your second equation. That should be a red flag, because you're multiplying by 0 if x = 3. In the original function there will be a discontinuity at x = 3, and one needs to keep this fact in mind when solving for an "inverse."
brotherbobby said:
My example above in this thread is a good one. There is nothing in the inverted function of ## = f^{-1}(y)## to suggest that the value ##y=5## is forbidden. And yet it is, separately seen from the function ##f(x)## itself.
The "inverted function" ##f^{-1}(y)## is not a function, since for nearly every value of y there are two values of x. Finding the inverse of a function is fraught with difficulties, even with polynomials of low enough degree that an inverse can be found. When you are dealing with rational functions, as in this thread, there are problems with restricted domain that need to be addressed.

Last edited:
PeroK

## 1. What is the possible value of the expression when x = 3?

When x = 3, the expression becomes undefined as it results in division by 0. Therefore, the expression has no possible value when x = 3.

## 2. Can the expression have a negative value?

Yes, the expression can have a negative value. For example, when x = -2, the expression becomes -4.

## 3. Is it possible for the expression to have a fraction as its value?

Yes, it is possible for the expression to have a fraction as its value. For example, when x = 6, the expression becomes 5/3.

## 4. What happens to the value of the expression as x approaches 3?

As x approaches 3, the expression approaches infinity. This is because as x gets closer to 3, the denominator becomes smaller and smaller, resulting in a larger value for the expression.

## 5. Can the expression have multiple possible values?

No, the expression can only have one possible value for each given value of x. This is because each value of x corresponds to a unique value for the expression, and there can only be one value for a given input.

• Precalculus Mathematics Homework Help
Replies
13
Views
388
• Precalculus Mathematics Homework Help
Replies
11
Views
618
• Precalculus Mathematics Homework Help
Replies
7
Views
481
• Precalculus Mathematics Homework Help
Replies
15
Views
729
• Precalculus Mathematics Homework Help
Replies
17
Views
220
• Precalculus Mathematics Homework Help
Replies
23
Views
692
• Precalculus Mathematics Homework Help
Replies
6
Views
112
• Precalculus Mathematics Homework Help
Replies
3
Views
360
• Precalculus Mathematics Homework Help
Replies
2
Views
542
• Precalculus Mathematics Homework Help
Replies
10
Views
910