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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 \[x=\ln \sqrt{y}\]
 11 months ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 11 months 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?
 11 months ago

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

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

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

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

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

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

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