f(x) = 63/4x-1 1

Question

f(x) = 63/4x-1       1<x<16
g(x) = xsquared     -4<x<-1
why, when solving fg(x)=1, do you reject one of the solutions???

rajee_sam · Answer

is it f o g(x)??

rajee_sam · Answer

or (f . g) (x) ??

rajee_sam · Answer

@nevarm ???

anonymous · Answer

f of g.

anonymous · Answer

sorry @rajee_sam lost connection for a moment, am back now :)

rajee_sam · Answer

$$g (x) = x ^{2} \space and \space f(x) = \frac{ 63 }{ 4x - 1 }$$ $$f(g(x)) = f( x ^{2}) =  \frac{ 63 }{ 4x ^{2}-1 }$$

anonymous · Answer

yeah, i got that too

rajee_sam · Answer

now when you solve this for = 1$$\frac{ 63 }{ 4x ^{2} - 1 } = 1 ; 63 = 4x ^{2} -1$$

rajee_sam · Answer

$$4x ^{2} - 64 = 0 ; 4(x ^{2} - 16) = 0$$

rajee_sam · Answer

now the solutions are x = -4 and x = +4

rajee_sam · Answer

but the domain for f(x) is 1<f(x) <16, since -4 does not belong here, we discard it

anonymous · Answer

but why do we not also consider the domain of g(x),                         {which is -4,g(x)<-1}

anonymous · Answer

?

rajee_sam · Answer

sorry 1<x<16

rajee_sam · Answer

you have absorbed the g(x) into f(x) it is only f(x) now

rajee_sam · Answer

if it had been (g o f )(x) then you would take -4<x<-1 domain and discard +4

rajee_sam · Answer

or you would discard both because -4 is not inclusive in the interval

anonymous · Answer

ahh. i see thanks, rajee :) x

anonymous · Answer

yeah, i think i get it now. thank you very much for explaining that to me x much appreciated