\[\dfrac{x^2+x-6}{x-2}=x+3\]
In the view of part (a), is this equation correct:
\[ lim_{x \rightarrow 2 } \space \dfrac{x^2 +x-6}{x-2} = lim_{x\rightarrow 2 }(x+3) \]
This problem aims to find the equation’s right domain, making it an equivalent fraction. The concepts required for this problem are related to quadratic algebra which includes domain, range interception, and undefined functions.
Now the domain of a function is the group of values that we are permitted to put into our function, where such group of values is represented by the x terms in a function such as f(x). Whereas the range of a function is a group of values that the function accepts. When we plug in the x values in that function, it shoots out the range of that function in the form of a group of values.
Expert Answer
We need to understand the value of domain because it helps to define a relationship with the range of the function.
Part a:
Let’s first factorize the left-hand side of the equation so it becomes easy to solve it:
\[=\dfrac{x^2 + x – 6}{x -2}\]
\[=\dfrac{x^2 + (3 – 2)x – 6}{x -2}\]
\[=\dfrac{x^2 + 3x – 2x – 6}{x -2}\]
\[=\dfrac{(x – 2)(x + 3)}{x -2}\]
So here we have a common factor $(x-2)$ which can be canceled out. Thus we have $(x+3)$ left on the left-hand side.
Note that we have simplified the left-hand side to be equal to the right-hand side of the equation. So if we plug $x = 2$ into the expression $x + 3$, we do not get an undefined value, which is okay. but doing the same for the expression $ \dfrac{x^2 + x-6}{x-2} $ gives us an undefined value.
This is because we would get a $0$ in the denominator, resulting in an undefined value.
Therefore we cannot say that:
\[\dfrac{x^2 + x – 6}{x -2}=x+3\]
Unless we make a requirement in the above expression that is:
\[x\neq 2\]
Our expression becomes:
\[\dfrac{x^2 + x – 6}{x -2}=x+3,\space x\neq 2\]
The above expression states that all numerical values are allowed as the domain of the function, with the exclusion of the value $2$ which explicitly result in an undefined value.
Part b:
Yes, the expression is correct since you can reach as close to $2$ as you desire and these functions will still be equal. At the actual value $x=2$, these $2$ functions become unequal as stated in part $a$.
Numerical Result
The domain must be mentioned with the expression, otherwise it will result in an undefined value.
\[\dfrac{x^2 + x – 6}{x-2}=x+3,\space x\neq 2\]
Example
What is wrong with this equation?
$\dfrac{x^2 + x – 42}{x-6}=x+7$
We understand that for a fraction to exist, the denominator must be a positive number and it should not be equal to $0$.
Since we don’t have variables on the right-hand denominator, $x+7$ is achievable for all values of $x$, whereas the left-hand side has a denominator of $x-6$. For $x-6$ to be a positive number:
\[x>6; x\neq 6\]
Thus, our expression becomes:
\[\dfrac{x^2 + x – 42}{x -6}=x + 7,\space x\neq 6\]