1. ## function

The function g is defined by g(x)=(x+1)/(x-2), x is not equal to k and m, find the values of k and m

of course, I know that x is definitely not equal to 2, but how about the other one?

2. ## Re: function

That's it -- x cannot equal 2. Are you sure you wrote the function correctly?

3. ## Re: function

The other value would be $\displaystyle k+m-2$.

4. ## Re: function

I suspect there is some aspect of this problem being ignored. Is this the exact and complete language of the problem?

$a \in \mathbb R\ and\ a \ne 0 \implies \dfrac{1}{a} \in \mathbb R.$

$\therefore a, b \in \mathbb R\ and\ a \ne 0 \implies b * \dfrac{1}{a} \equiv \dfrac{b}{a} \in \mathbb R.$

So if the only restriction on g(x) is that it is a real-valued function, x = 2 is the only number necessarily outside its domain. Of course the definition of g may exclude any other real number.

5. ## Re: function

Perhaps the "other value" is for $\displaystyle g(x)$. While $\displaystyle x \neq 2$, it is fairly straightforward to show that $\displaystyle g(x) \neq 1$ for any value of $\displaystyle x$. In other words, $\displaystyle g(x)$ has an inverse function for any value except at $\displaystyle g(x)=1$.

6. ## Re: function

Originally Posted by SlipEternal
Perhaps the "other value" is for $\displaystyle g(x)$. While $\displaystyle x \neq 2$, it is fairly straightforward to show that $\displaystyle g(x) \neq 1$ for any value of $\displaystyle x$. In other words, $\displaystyle g(x)$ has an inverse function for any value except at $\displaystyle g(x)=1$.
That's a very clever guess.

7. ## Re: function

Originally Posted by JeffM
I suspect there is some aspect of this problem being ignored. Is this the exact and complete language of the problem?

$a \in \mathbb R\ and\ a \ne 0 \implies \dfrac{1}{a} \in \mathbb R.$

$\therefore a, b \in \mathbb R\ and\ a \ne 0 \implies b * \dfrac{1}{a} \equiv \dfrac{b}{a} \in \mathbb R.$

So if the only restriction on g(x) is that it is a real-valued function, x = 2 is the only number necessarily outside its domain. Of course the definition of g may exclude any other real number.
yeah, that's it. The answer given is k=2, m=5. I can't understand

8. ## Re: function

Originally Posted by Trefoil2727
yeah, that's it. The answer given is k=2, m=5. I can't understand
Given the answer, I can come up with a question that would have that as its answer. What values of $\displaystyle x$ are not in the domain of $\displaystyle (g\circ g)$? That is the function $\displaystyle g$ composed with itself.

$\displaystyle (g\circ g)(x) = g(g(x)) = \dfrac{g(x)+1}{g(x)-2} = \dfrac{\dfrac{x+1}{x-2}+1}{\dfrac{x+1}{x-2}-2}$

From here, you can simplify:

$\displaystyle \dfrac{\dfrac{x+1}{x-2}+1}{\dfrac{x+1}{x-2}-2}\cdot \dfrac{x-2}{x-2} = \dfrac{x+1+x-2}{x+1-2(x-2)} = \dfrac{2x-1}{5-x}$

Obviously, $\displaystyle x\neq 5$ for this. But, since $\displaystyle (g\circ g)(x)$ is only defined when $\displaystyle g(x)$ is defined, we also have $\displaystyle x \neq 2$.

9. ## Re: function

Clearly $\displaystyle f(x)= \frac{x+1}{x- 2}$ is undefined when x= 2. To determine a value that f(x) cannot be equal to, do the division: $\displaystyle \frac{x+1}{x- 2}= 1+ \frac{3}{x- 2}$. Since the fraction is never 0, f(x) is never 1.