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

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: nipuna92 on May 10, 2010, 12:53:44 pm

Title: P1 Maths October November 2005 question 8
Post by: nipuna92 on May 10, 2010, 12:53:44 pm

Can someone answer only part 2 of (ii)

A function f is defined by f : x ? (2x ? 3)3 ? 8, for 2 ? x ? 4.

(i) Find an expression, in terms of x, for f(x) and show that f is an increasing function. [4]
(ii) Find an expression, in terms of x, for f?1(x) and find the domain of f?1. [4]

this is the inverse function of f(x)

(cuberoot of x+8/2) + 1.5
Can someone find the domian,please include all ur steps and can u find the range of f(x) what i mean is like is there a way to find it...so that then finding the domain of f inverse is easy

thanx in advance

Title: Re: P1 Maths October November 2005 question 8
Post by: nid404 on May 10, 2010, 01:26:22 pm

Nov 05 Q8)

i) y= (2x-3)³-8

y+8=(2x-3)³

Title: Re: P1 Maths October November 2005 question 8
Post by: nipuna92 on May 10, 2010, 01:28:30 pm

I need the way to find the domain not the inverse function
thanx anyway for helping
and also can the range for f(x) be found

Title: Re: P1 Maths October November 2005 question 8
Post by: nid404 on May 10, 2010, 01:39:10 pm

when you find the inverse. The domain of the original function becomes the range for the inverse function. In the same way the range of the original function is the same as the domain of the inverse function.

Range of the f(x) can be found by substituting the limits  

when x=2
f(x)=-7
 
when x=4
f(x)=117

this becomes the domain of the inverse

so domain of the inverse is

-7<= x <=117

Title: Re: P1 Maths October November 2005 question 8
Post by: nipuna92 on May 10, 2010, 01:51:14 pm

thanz u have been clearing my doubts all day long.hehe ;D
best of luck to u as well for the exam. ;)

Title: Re: P1 Maths October November 2005 question 8
Post by: nid404 on May 10, 2010, 01:54:19 pm

No problem :)

And Thank you  ;D ;D