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We have given
It is linear function so its inverse will exist.
An algebraic rule, set that functional product of one to one function is again a one to one function.
Indeed: `f: A---> B` ; `g:B--->C` so:
`AAy inB ---> EEx inA | y=f(x) `
`AA z in C ---> EE y in B | z=g(y)` (Surriettive property)
`f(x)=f(x_0) rArr x= x_0` `g(y)=g(y_0) rArr y=y_0` (innietive rules)
So if `z=z_0,EE y,y in A_0 | g(y) = z, g(y_0)= z_0` (1)
For suriretvie property of `g(y)`
Further `EE x. x_0 in A | f(x)=y; f(x_0)= y_0` (2)
Cause inniective rules (1) reports: `y=y_0` , and agian by (2)
Now in the function we have to find is product of
`f: RR ----> RR`
`f^(-1): RR-----> RR`
So the function product `f@ f^(-1)` (That generally doesn't collides with identity) is product of `y=f(x)= 3x +1` and it's inverse, `x=f^(-1)(y)`,that, being both linear, are one to one functions.
So that, their product is again a one to one function, as we have shown above, that means `x_0 in RR` is image only and only of itself.
In conclusion we get: `f@ f^(-1) (5)=5;f@f(-1) (6)= 6` ``
at relation (1) :
"So if `z=z_0,EE y,y inA_0|g(y)=z,g(y_0)=z_0` (1) "
You should read instead:
" So if `z=z_0,EE y,y_0 in A| g(y)=z,g(y_0)=z_0` (1)"
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