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Suppose a function is defined as $ f(x) =\frac{x^2 - 9}{x-3}. $ If we divide the common factor $ x-3 $ from both the numerator and denominator :- $$ \frac{x^2 - 9}{x - 3} \\ = \frac{(x+3)(x-3)}{x-3} \\ = x+3 $$

1) Is $ x+3 $ a different function from the original $ \frac{x^2 - 9}{x - 3} $ ?

2) Since $ \frac{x^2 - 9}{x - 3} $ does not have 3 in its domain, when we simplify it to become $x + 3$, does it now have 3 in its domain ?

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Two functions are equal if and only if they have the exact same domain, and the exact value at each element of the domain. (If you like your functions to also have a specified codomain, then you also require the functions to have the exact same codomain).

“Cancelling” a common factor includes the implicit assertion that this common factor is not equal to $0$. So in your example, when you cancel $x-3$, you should also be saying under your breath “and $x\neq 3$”, so the correct simplification would be that $\frac{x^2-9}{x-3}$ is equal to the function $x+3,\ x\neq 3$.

(So, yes, the function $f(x) = \frac{x^2-9}{x-3}$ and the function $g(x) = x+3$ are different functions, because they have different domain; on the other hand, $f(x)$ is equal to the function $h(x) = x+3$, $x\neq 3$. )

However, cancelling does not always lead to a different function. For example, the function $$f(x) = \frac{x^2-5x+6}{x^2-6x+9}= \frac{(x-3)(x-2)}{(x-3)^2}$$ is equal to the function $$g(x) = \frac{x-2}{x-3}$$ because they have the exact same domain (all $x\neq 3$), and take the same value at every point of the domain.

(This is similar to the fact that the function $$f(x) = \frac{1}{\quad\frac{1}{x}\quad}$$ and the function $g(x) = x$ are different functions, because they have different domains.)

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$x+3$ is different from $\frac{x^2-9}{x-3}$, but only in the domain. All the simplification does is to add $x=3$ to the domain.

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  • $\begingroup$ After the simplification, does the domain of the function 'f' include 3 ? $\endgroup$ – arandomguy May 25 '19 at 6:02
  • $\begingroup$ After the cancellation, you have a different function. The function that results from cancelling does include $3$ in the domain; and that’s why it is different from the function $f$ that you started with. $\endgroup$ – Arturo Magidin May 25 '19 at 6:04
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At the very first we should understand what is the function is.

A function is defined as,

let, $X$ and $Y$ are respectively called the domain and the codomain of the function $f$. If $(x, y)$ belongs to the set defining $f$, then $y$ is the image of $x$ under $f$, or the value of $f$ applied to the argument $x$. Especially in the context of numbers, one says also that $y$ is the value of $f$ for the value $x$ of its variable, or, still shorter, y is the value of $f$ of $x$, denoted as $y = f(x)$.

In your question, you haven't defined a domain and range we can't give an exact answer. If your domain and range of both equations are the same, we can say that both functions are the same. But if your domain and range are different with each equation, the functions are not the same.

For further study, search real-analysis definitions of function, domain, co-domain, and range.

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