Question

Let f, g and h are functions defined by f(x) = x-1, g(x) = ${{x}^{2}}-2$ and h(x) = ${{x}^{3}}-3$. Show that (fog)oh = fo(goh)

Answer 1

Hint:For solving this problem, we should be aware about the concepts of composite functions. Thus, (fog) (x) = f(g(x)). Basically, the function g(x) acts as the input to f(x) in this case. We make use of this to prove in the case of the above problem.

Complete step by step answer:
We first try to understand the meaning of composite functions briefly. Basically, a composite function is a function that depends on another function. A composite function is created when one function is substituted into another function. Thus, to explain, we take an example. Suppose, we are given that f(x) = x+1 and g(x) = 1 –x, then to find the composite function f (g (x)), we have,
f(1-x) = (1-x) + 1 = 2-x
(That is, we basically substitute g(x) in place of x in the function f(x).)
Now, coming back to the problem in hand, we have,
f(x) = x-1, g(x) = ${{x}^{2}}-2$ and h(x) = ${{x}^{3}}-3$
On LHS, we have,
= (fog)oh
= (f(g(x))oh
= (f(${{x}^{2}}-2$))oh
= $({{x}^{2}}-2-1)$ oh
= $({{x}^{2}}-3)$oh
Let, a(x) =$({{x}^{2}}-3)$. So, we have,
= a(h(x))
= $({{({{x}^{3}}-3)}^{2}}-3)$
=$({{x}^{6}}+9-6{{x}^{3}}-3)$
= ${{x}^{6}}-6{{x}^{3}}+6$
Now, solving RHS, we have,
= fo(goh)
= fo(g(h(x))
= fo(g (${{x}^{3}}-3$))
= fo$({{({{x}^{3}}-3)}^{2}}-2)$
= fo$({{x}^{6}}+9-6{{x}^{3}}-2)$
= fo(${{x}^{6}}-6{{x}^{3}}+7$)
Let, a(x) =${{x}^{6}}-6{{x}^{3}}+7$, thus,
= f(a(x))
= f (${{x}^{6}}-6{{x}^{3}}+7$)
= ${{x}^{6}}-6{{x}^{3}}+7$-1
= ${{x}^{6}}-6{{x}^{3}}+6$
Thus, LHS=RHS, hence, proved.

Note: While solving the problems related to composite functions, one should always be careful about the domain constraints associated while solving each step. For example, if we have, f(x) = $\dfrac{1}{1-x}$ and g(x) = 1. Then, in this case f(g(x)) would not exist since this would make the denominator 0 and which is not possible. However, g(f(x)) exists in this case.