IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: mz on May 10, 2010, 02:56:30 pm

Title: Math P1 s07
Post by: mz on May 10, 2010, 02:56:30 pm

can anybody please explain the concept behind question number 11, part ii?

Title: Re: Math P1 s07
Post by: immortal on May 10, 2010, 03:09:54 pm

Find da inverse of f(x) dat is f^-1(x)
Domain of f^-1(x)=range of f(x) & vice-versa

Title: Re: Math P1 s07
Post by: nid404 on May 10, 2010, 03:26:24 pm

Quote from: mz on May 10, 2010, 02:56:30 pm

can anybody please explain the concept behind question number 11, part ii?

here u go


Let f-1(x) = y
i.e. f(y) = x

x= 6/ 2y+3

2y+3=6/x
2y= 6/x -3
y= (6-3x)/2

thus f^-1(x)= (6-3x)/2
y >= 0 implies that 6-3x >= 0 i.e.
6 >= 3x  i.e.  x <= 2
Thus the domain of f-1(x) is 0 <=  x  <= 2

Title: Re: Math P1 s07
Post by: mz on May 10, 2010, 03:40:08 pm

thankyou so much both of you!!!!