Question

Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.

please answer this​

Answer 1

Answer:

$let \: f(x1) = f(x2) \\ 4x1 + 3 = 4x2 + 3 \\ 4x1 = 4x2 \\ x1= x2 \\ hence \: f \: is \: one \: one \: \\ let \: y \: = f(x) \\ y = 4x + 3 \\4 x = y - 3 \\ x = \frac{y - 3}{4} \\ now \: f( \frac{y - 3}{4} ) = 4( \frac{y - 3}{4} ) + 3 \\ = \frac{4y - 3 + 3}{4} = y \\ hence \: f \: is \: onto \: \\ since \: f(x) \: is \: bijective \: ie \: one \: one \: onto \: hence \: {f}^{ - 1} \: exists \: \: \: \\ {f}^{ - 1} = \frac{y - 3}{4}$

Answer 2

the given function,N-->Y,is defined as f(x)=4x+3

where,y={y€N:y=4x+3 for sum x€N}

now, f(x)=4x+3

f'(x)=4>0

f is a strictly increasing function

f is one-one

also here y=(7,8,9....... infinity)

and range of f=(7,8,9..... infinity)

range of f=y=codomain of f

=> f is onto

=> f is invertible

now let's,Y€y such that f(x)=y

x=f^-1y

x€N

y=4x+3=>x=(y-3)/4

from here we get

inversw of f, f^-1y=(y-3)/4

this is the result which we get from here.....

{hope it helps}