C£ very quick range check plz 2secs! Watch

skdoc
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#1
Report Thread starter 12 years ago
#1
if f(x)= 1/x + 5 what is the range?


I am a bit confused because x=0 is not mapped ?? anyone?
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spursrule
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#2
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#2
xER bcoz it can take any vaues
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skdoc
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#3
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#3
but x can't take 0 value can it ?
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wacabac
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#4
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#4
f(x) takes all values in R except it is asymptotic at f(x)=5 (f(x) tends to, but never actually equals 5)
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abood
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#5
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#5
gotta say x=/ (not equal to) 0 then state the xER
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wacabac
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#6
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(Original post by skdoc)
but x can't take 0 value can it ?
If x were to take the value 0 then it would be undefined as, 1/0 is undefined. But as x->0 ie as x gets really, really small, then f(x) gets bigger and bigger.
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wacabac
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#7
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(Original post by aboodz)
gotta say x=/ (not equal to) 0 then state the xER
But we're talking about the range, not the domain.
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skdoc
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#8
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#8
so the range wouldbe ..... xeR x notequalsign 0 ?
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wacabac
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#9
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#9
The range is the set of values that f(x) can take not x. So the range would be f(x)eR, f(x) =/= 5
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bluenoxid
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#10
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#10
So x is greater than 0 ?

Sorry edit 5
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skdoc
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#11
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#11
gotcha thx! to poster above yes all y values are greater than 5 but also all less than 5 BUT not equal to 5 ... hence we say the function range is xeR all real values of yaxis EXCEPT y not= 5 ! hope that helps
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__WiZaRD__
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#12
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#12
ok so whats the range?

f(x) e R , f(x) <> 5 ?
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skdoc
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#13
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#13
yes is <> means not equal thats correct range
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wacabac
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#14
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#14
(Original post by skdoc)
the function range is xeR
No, because x is the domain. The function range is f(x)eR, f(x) =/= 5 (because no value of x will give f(x)=5). The function range is not xeR, that is the domain.

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