Let f : R → R : f(x) (2x - 3) and \(g:R→R:g(x)=\frac{1}{2}(x+3)\).

1.	Let f : R → R : f(x) (2x - 3) and \(g:R→R:g(x)=\frac{1}{2}(x+3)\). Show that (f o g) = IR = (g o f).
Answer» To prove: (f o g) = I_R = (g o f). Formula used: (i) f o g = f(g(x)) (ii) g o f = g(f(x)) Given: (i) f : R → R : f(x) = (2x - 3) (ii) \(g:R→R:g(x)=\frac{1}{2}(x+3)\) Solution: We have, f o g = f(g(x)) \(=f(\frac{1}{2}(x+3))\) \(=[2(\frac{1}{2}(x+3))-3]\) = x + 3 – 3 = x = I_R g o f = g(f(x)) = g(2x - 3) \(=\frac{1}{2}(2x-3+3)\) \(=\frac{1}{2}(2x)\) = x = I_R Clearly we can see that (f o g) = IR = (g o f) = x Hence Proved.

Let f : R → R : f(x) (2x - 3) and \(g:R→R:g(x)=\frac{1}{2}(x+3)\). Show that (f o g) = IR = (g o f).

Answer»

To prove: (f o g) = I_R = (g o f).

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = (2x - 3)

(ii) \(g:R→R:g(x)=\frac{1}{2}(x+3)\)

Solution: We have,

f o g = f(g(x))

\(=f(\frac{1}{2}(x+3))\)

\(=[2(\frac{1}{2}(x+3))-3]\)

= x + 3 – 3

= x

= I_R

g o f = g(f(x))

= g(2x - 3)

\(=\frac{1}{2}(2x-3+3)\)

\(=\frac{1}{2}(2x)\)

= x

= I_R

Clearly we can see that (f o g) = IR = (g o f) = x

Hence Proved.