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anonymous
 one year ago
f(x)=ln(x)+e^(1/x1). What´s the domain for f?
Any suggestions on where to start? :)
anonymous
 one year ago
f(x)=ln(x)+e^(1/x1). What´s the domain for f? Any suggestions on where to start? :)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since the function f(x) has two distinct parts, you should start with them and then find the intersection of the two individual domains, like for Ln(x) the domain would be_

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1You have two things to worry about \(x>0\) because of the domain of \(\ln(x)\) and \(x1\ne 0\) because of the \(\dfrac{1}{x1}\) (we cant divide by \(0\)). So we can use all positive numbers except \(1\). Domain is \(\{x\mid x>0\text{ and } x\ne 1\}=(0,1)\cup (1,\infty)\) Make sense?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0hmm, might be looking at it in a mirror tho ... the union of the bad parts that is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it makes totally sense! Thanks!

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I think we want the intersection of the two @amistre64

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Of course XD Thought it was a math term that I had missed :D

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0intersection of the domains yes ... i was thinking about the exclusions at first. we want to make sure all the bad parts are avoided for each term

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0if x=2,3 are bad for the first part, and x=2,4,5 are bad for the last part; then we would exclude x=2,3,4,5 was in my head

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0time to build a house ... have fun
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