Qualification > Math
Additional Math Help HERE ONLY...!
cooldude:
--- Quote from: Dibss on June 05, 2010, 01:05:16 pm ---5 A function f is defined by for the domain x 0.
(i) Evaluate f^2(0). [3]
(ii) Obtain an expression for f^-1. [2]
(iii) State the domain and the range of f^-1. [2]
I don't understand how to get the correct answer for part (iii) so could someone please help me?
Mark scheme attached.
--- End quote ---
first of all the range of f^-1(x)=domain of f(x) and range of f(x)=domain of f^-1(x)
this will answer the part of the q asking the range of f^-1 wich is f^-1(x)=>0
now we use the part that range of f(x)=domain of f^-1(x) , when we subsitute 0 in (e^x)+1/4 we get 1/2, we substitute 0 as this is the minimum value in the domain given, so thus we get the range as f(x)=>0.5 and this is equal to the domain of f^-1(x), i.e. x=>0.5
Dibss:
--- Quote from: cooldude on June 05, 2010, 01:26:35 pm ---first of all the range of f^-1(x)=domain of f(x) and range of f(x)=domain of f^-1(x)
this will answer the part of the q asking the range of f^-1 wich is f^-1(x)=>0
now we use the part that range of f(x)=domain of f^-1(x) , when we subsitute 0 in (e^x)+1/4 we get 1/2, we substitute 0 as this is the minimum value in the domain given, so thus we get the range as f(x)=>0.5 and this is equal to the domain of f^-1(x), i.e. x=>0.5
--- End quote ---
Oh, i get it!
Thank you =]
+ REP
cooldude:
--- Quote from: Dibss on June 05, 2010, 01:31:45 pm ---Oh, i get it!
Thank you =]
+ REP
--- End quote ---
np and Thanks for the rep ;D
jellybeans:
:o Help pleaaaase! 5ii) THANK YOU. :P :P :P :P
J.Darren:
http://img198.imageshack.us/img198/135/velocityquestion2.png
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version