QSHYRN'S QUESTION threAD

[qshyrn]

let a be a +ve no. let f: [2,infinity) --> R, f(x)=a-x  and let  g:(neginfinity,1] --> R, g(x)= x^2 +a. Fine all values of 'a' for which  f(g(x)) and g(f(x)) both exist..   can someone explain how to solve this.   Thanks

[TrueTears]

,  

So 1. we need to find so that any value of will only be a subset of

The function is a straight line and always decreasing, so the maximum value will be at its left endpoint, at . Hence, we can say . For it to have the correct range, we require

and 2. we need to find so that any value of will only be a subset of

For , determines the vertical translation of the standard parabola with vertex at (0,0). We can say . So to have the right range, .

Hence the answer is .

[qshyrn]

,   

So 1. we need to find so that any value of will only be a subset of

The function is a straight line and always decreasing, so the maximum value will be at its left endpoint, at . Hence, we can say . For it to have the correct range, we require

and 2. we need to find so that any value of will only be a subset of

For , determines the vertical translation of the standard parabola with vertex at (0,0). We can say . So to have the right range, .

Hence the answer is .

wow thanks .. too bloody quick

[qshyrn]

does anyone have an explanation  type proof to why  f(f^-1(x)) is x ??

[d0minicz]

let



so
= x

[TrueTears]

let



so
= x

But can you generalise?

[d0minicz]

nope not me

[TrueTears]

If we define some function

then by definition its inverse function is

Thus

[qshyrn]

If we define some function

then by definition by definition its inverse function is

Thus

thx thikni sorta get it now. the essential book didnt explain it  clearly..

[addikaye03]

f(x) and f^-1(x) are mutually inverse functions where their graphs are reflective over the line y=x, so obviously any solutions cancel and follow the line y=x.

So consider

f:f(x)= 3x

f^-1: x=3y

therefore inverse is y=3/x

f(f^-1(x))=x