Question

I've been asked to determine the Range of this function:
f(x)= Sqrt(1-x)/(x^2-1)
I have done the usual algebra but ended up with a cubic equation in x and y that I don´t know how to solve. Can anyone help?

Answer 1

What usual algebra did you do? Can y = 0?

Answer 2

Thanks, tkhunny.
I tried to isolate the "x" in y=Sqrt(1-x)/(x^2-1), obtaining:
\[x ^{3}+x ^{2}-x+\frac{ 1 }{ y ^{2} }-1=0\]

I think that to get the range of the original function, I need to solve the above for x as a function of "y", but I couldn´t.

Answer 3

I see. Not really a good plan. It might be if you could do that, but as you are suspecting, it cannot always be done. You just have to reason it out a piece at a time.

\(\sqrt{x-1}\) is simple enough. You should see the range as \(0 \le \sqrt{x-1} \le \infty\)

Similarly, \(x^{2] - 1\) is simple enough. You should see the range as \(-1 \le \sqrt{x-1} \le \infty\)

Now the tricky part...Let's just start from the right.

For \(x > 1\), both numerator and denominator are positive, so we'll nor be wandering either to zero or to negative values. The denominator grows far faster than the numerator, so this must approach zero as x increases without bound.

For \(x = 1\), both numerator and denominator are zero, so \(x = 1\) isn't even in the Domain. An examination of the behavior near 1, on the positive side shows the necessary behavior.

f(2) = 0.3333
f(1.5) = 0.5666
f(1.1) = 1.5058
f(1.01) = 4.9751
f(1.001) = 15.8034
f(1.0001) = 49.9975

See where this is going?

So far, for \(1 \lt x \lt \infty\), we have the Range: \(0 \lt y \lt \infty\)

For x < 1, the numerator fails to exist, so we are done.

It's not always obvious how to proceed. I realize this may be a lot to take in. Welcome to the world of abstraction!