Since we are given isn't in the domain, we know that any number of the form isn't in the range of f.
1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)
2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.
First time seeing this. Somehow tell me how to approach it, I don't need the answer.
has a vertical asymptote at and a horizontal asymptote at . Therefore, has a horizontal asymptote at and a vertical asymptote at .
Thus so that .
Also, the zero of is equal to the y-intercept of , and vice-versa, but I see that this just gives again.
So, we have that .
We also know that for Solve and that should give us
Therefore, Cancel .
Duh. Thanks. I wasn't thinking straight. If , then is an even function! But this isn't always true for this function: we're just looking for values of x where this would be true.
So this brings me to another question: how do we know that the function at issue here is an odd function?
Note that being odd always implies that (you don't need the other condition). Indeed, if for all , then substituting a in for yields . But whence . Adding to both sided yields . Finally, divide both sides by to get .
Edit: I should clarify that being odd always implies that whenever is in the domain of .