• Nov 20th 2006, 06:52 PM
lisaak
Here's the question:
Let R be the set of real numbers and let f be the function from R to R defined by
f(x) = 1/[(x^2)+1] for all x in R
Let C= {r in R|r<(1/5)}.
Find the set f^-1 (C)

From what I understand, I get:
f^-1(C)={x in R|[1/[(x^2)+1]] < (1/5)
I'm not sure if that is even close to being correct, but after I get to this point, I have no clue where to go with it.
• Nov 20th 2006, 06:59 PM
ThePerfectHacker
Quote:

Originally Posted by lisaak
Here's the question:
Let R be the set of real numbers and let f be the function from R to R defined by
f(x) = 1/[(x^2)+1] for all x in R
Let C= {r in R|r<(1/5)}.
Find the set f^-1 (C)

Watch frum die meister.

$\displaystyle f:\mathbb{R}\to \mathbb{R}$
Und,
$\displaystyle f(x)=\frac{1}{1+x^2}$.
We have,
$\displaystyle C=(-\infty,1/5)$
Thus,
$\displaystyle f^{-1}[C]=\{x\in \mathbb{R}|f(x)\in (-\infty,1/5)\}$
That means, find all $\displaystyle x\in \mathbb{R}$
Such as,
$\displaystyle \frac{1}{1+x^2}<1/5$
Since both sides are positive you can take reciprocals and flip the signs (a trick not taught in schools)
$\displaystyle 1+x^2>5$
Equivalently,
$\displaystyle x^2>4$
Equivalently,
$\displaystyle |x|<2$
Thus,
$\displaystyle f^{-1}[C]=(-2,2)$