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osanseviero Group TitleBest ResponseYou've already chosen the best response.1
So they want the inverse of \[F(x)=\ln \sqrt{x}\] So the inverse is \[x=\ln \sqrt{y}\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yep, now isolate y
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\Large a=\ln c \implies c = e^a \)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
\[y=\left( e ^{x} \right)^{2}\] the 2 can go down, right?
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
y=2e^x
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
not actually
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\Large (e^x)^2 = e^{2x}\)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
oops, another silly mistake
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
Can someone help me demostrate that (FoFinverse)(x)=(FinverseoF)(x)=x ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
demonstrate ? you have a function for verifying that ?
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
The one we just did
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
\[\ln \sqrt{e ^{2x}} = e ^{2\ln \sqrt{x}}=x\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
so, f inverse = e^2x just plug in x =ln \sqrt x oh, you did it :)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
But how and why? (I think that was a good statement)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
And how would that be x?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
ln x^m = m ln x
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\sqrt {(e^x)^2} = e^x \\ \ln e^x = x \ln e = x\)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
so the left one is kinda obvious, so it is x
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
Oh...and the other one is with \[a ^{logaN}=\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\huge a^{\log_aN}=N\)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
:) Now I understand, thanks for fifth time, just one more question :P
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
Which would be the domain and the range of the first, normal function?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
domain of ln sqrt x ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
since its sqrt x, x>= 0 since its ln, sqrt x >0 in all, x>0
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
so the range of the inverse is x>0
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
Wish me good luck tomorrow
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes! best of luck! hope you get full marks :)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
I hope so, this will be a hard examn...all types of functions, trigonometric identities, operations with functions
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
if you have practices enough, nothing is hard :)
 one year ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
Thanks for the help all night!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you're welcome ^_^
 one year ago
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