Question -
Answer -
f(x) and g(x) arepolynomials.
⇒f: R → R and g: R → R.
So, fog: R → R and gof: R → R.
(i) (fog) (x) = f (g(x))
= f (x2 +1)
= 2 (x2 +1) + 5
=2x2 +2 + 5
= 2x2 +7
(ii) (gof) (x) = g (f(x))
= g (2x +5)
= (2x + 5)2 +1
= 4x2 +20x + 26
(iii) (fof) (x) = f (f(x))
= f (2x +5)
= 2 (2x + 5) + 5
= 4x + 10 + 5
= 4x + 15
(iv) f2 (x)= f (x) x f (x)
= (2x + 5) (2x +5)
= (2x + 5)2
= 4x2 +20x +25
Hence, from (iii) and(iv) clearly fof ≠ f2