Let f : R → R : f(x) = x2 and g : R → R : g(x) = (x + 1). Show th

1.	Let f : R → R : f(x) = x2 and g : R → R : g(x) = (x + 1). Show that (g o f) ≠ (f o g).
Answer» To prove: (g o f) ≠ (f o g) Formula used: (i) g o f = g(f(x)) (ii) f o g = f(g(x)) Given: (i) f : R → R : f(x) = x² (ii) g : R → R : g(x) = (x + 1) Proof: We have, g o f = g(f(x)) = g(x²) = ( x² + 1 ) f o g = f(g(x)) = g(x+1) = [ (x+1)² + 1 ] = x² + 2x + 2 From the above two equation we can say that (g o f) ≠ (f o g) Hence Proved

Let f : R → R : f(x) = x2 and g : R → R : g(x) = (x + 1). Show that (g o f) ≠ (f o g).

Answer»

To prove: (g o f) ≠ (f o g)

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

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

Given: (i) f : R → R : f(x) = x²

(ii) g : R → R : g(x) = (x + 1)

Proof: We have,

g o f = g(f(x)) = g(x²) = ( x² + 1 )

f o g = f(g(x)) = g(x+1) = [ (x+1)² + 1 ] = x² + 2x + 2

From the above two equation we can say that (g o f) ≠ (f o g)

Hence Proved