help! http://prntscr.com/1at5k0
do u know how to find the inver4se function?
the range of a function is the domain of the inverse
i knew it but i cannot remember it clearly now can you tell
@julian25
what's the range of a function?
the y values
right, so let's plug a few values in "x", and let's see what "y" becomes
ok..
let's say x = 10, 1,000, 1,000,000 so y becomes \(\cfrac{1}{9},\cfrac{1}{999},\cfrac{1}{999,999}\)
yeah
1/999,999 is really LESS than 1
ahan...
let's say x = -10, -1,000, -1,000,000 y becomes \(-\cfrac{1}{11},-\cfrac{1}{1,001},-\cfrac{1}{1,000,001} \)
so, - 1/1,000,001 is really LESS than -1
both of those values are a fraction, not 0, so, they're NOT 0, but LESS than -1 or 1
and notice that the "numerator" never changed
so, what do you think is the range from the choices given?
@jdoe0001 A ?
well, it says all real numbers, but 0 or \(\pm \infty\)
well, -10 is NOT 0, but is a real number, and "y" never became -10, or 10 or even 2 or -2
i am not getting anything :(
well, let's read B) it says the set of all rational number BUT 0 let's read C) all rational fractions \(\large \frac{p}{q}\) where is always p=1, and q is never 0 let's read D) all possible rational numbers so, which one do you think?
@jdoe0001 i think C
hmmmm, what makes "C" so attractive?
let's read it again
ok
\(\large \frac{p}{q}\) where "p" is always 1, so really \(\large \frac{1}{q}\) and "q" is never 0 well, originally it said that "x" was the "set of all integers BUT 1" if "x" becomes 1, "q" turns to 0 that is "x-1" x= 0 "1-1 = 0"
so, with THAT RESTRICTION that "x" is "the set of all integers BUT 1" pretty much guarantees that "q" in \(\frac{p}{q} \) NEVER becomes 0
yes
so, that's your range, a bunch of fractions, which are NEVER -1 OR 1 ever closing in to 0 but never getting to 0 :)
so the answer is C?
from the restrictions and notation, yes
ah finally! thanks a ton :)
yw
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