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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\Large a=\ln c \implies c = e^a \)
 9 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?
 9 months ago

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

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

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

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

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

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

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

osanseviero Group TitleBest ResponseYou've already chosen the best response.1
\[\ln \sqrt{e ^{2x}} = e ^{2\ln \sqrt{x}}=x\]
 9 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 :)
 9 months ago

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

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

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

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\huge a^{\log_aN}=N\)
 9 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
 9 months ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 9 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?
 9 months ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
domain of ln sqrt x ?
 9 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
 9 months ago

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes! best of luck! hope you get full marks :)
 9 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
 9 months ago

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

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

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