# If f(x) = \frac{x^{2} - 9x}{x - 9} and g(x) = x, is it true that f = g?

## Question:

If {eq}\displaystyle\; f(x) = \frac{x^{2} - 9x}{x - 9}\; {/eq} and {eq}\displaystyle\; g(x) = x {/eq},{eq}\; {/eq} is it true that {eq}\,f = g {/eq}?

## Domain of a Rational Function:

In mathematics, the domain of a function is the set of numbers that we can plug into that function and still have a defined function. A rational function is a function of the form {eq}\frac{f(x)}{g(x)} {/eq}. Since a denominator can never be zero, any number that makes the denominator of a rational function equal zero must be restricted from its domain.

We are given that {eq}\displaystyle f(x)=\frac{x^{2}-9x}{x-9} {/eq} and that {eq}g(x)=x {/eq}. Since {eq}\displaystyle f(x)=\frac{x^{2}-9x}{x-9} {/eq}, it is a rational function that can be simplified as follows:

• {eq}\displaystyle f(x)=\frac{x^{2}-9x}{x-9}=\frac{x(x-9)}{(x-9)}=x {/eq}

Thus, f simplifies to {eq}f(x)=x {/eq}, and it appears as though f = g. However, the original function of f is a rational function, so before doing any simplification, we must make any restrictions on the domain necessary. Any number that makes the denominator of f(x) equal 0 would need to be restricted from the domain of f. The denominator of f(x) is x - 9, and if we plug 9 into this, then we get 0, so the domain of f is all real numbers except for 9. Thus, we have the following:

• {eq}f(x)=x {/eq}, where x is all real numbers except for 9.

On the other hand, g(x) starts as {eq}g(x)=x {/eq}, so it is not a rational function, and there are no restrictions on the domain, so we have the following:

• {eq}g(x)=x {/eq}, where x can be any real number.

We see that though {eq}f(x)=x {/eq} and {eq}g(x)=x {/eq}, they aren't technically the same, because they have different domains.