Question

If \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R} \to \mathbb{R} \) be functions defined by \( f(x) = \cos x \) and \( g(x) = 3x^2 \) respectively, then prove that \( gof \neq fog \).

Hint:

Composition of functions is generally not commutative, meaning the order in which you apply the functions matters. Always calculate \(g(f(x))\) and \(f(g(x))\) separately to check for equality. A single counterexample (testing a specific value of \(x\)) is sufficient to prove they are not equal.

Answer 1

Step 1: Understanding the Concept:
This question requires an understanding of composite functions.
A composite function is created when one function is applied to the result of another function.
We need to find the expressions for \( gof \) (g-compose-f) and \( fog \) (f-compose-g) and then show that they are not identical.
Step 2: Key Formula or Approach:
The definitions of composite functions are:
1. \( (gof)(x) = g(f(x)) \)
2. \( (fog)(x) = f(g(x)) \)
To prove they are not equal, we need to show that \( g(f(x)) \neq f(g(x)) \) for at least one value of \(x\), or that their general expressions are different.
Step 3: Detailed Explanation or Calculation:
First, let's find the expression for \( (gof)(x) \).
By definition, \( (gof)(x) = g(f(x)) \).
We are given \( f(x) = \cos x \).
So, we substitute \( f(x) \) into \( g(x) \):
\[ (gof)(x) = g(\cos x) \] The function \( g(x) \) is defined as \( g(x) = 3x^2 \). To find \( g(\cos x) \), we replace \( x \) with \( \cos x \):
\[ (gof)(x) = 3(\cos x)^2 = 3\cos^2 x \] Next, let's find the expression for \( (fog)(x) \).
By definition, \( (fog)(x) = f(g(x)) \).
We are given \( g(x) = 3x^2 \).
So, we substitute \( g(x) \) into \( f(x) \):
\[ (fog)(x) = f(3x^2) \] The function \( f(x) \) is defined as \( f(x) = \cos x \). To find \( f(3x^2) \), we replace \( x \) with \( 3x^2 \):
\[ (fog)(x) = \cos(3x^2) \] Finally, let's compare \( (gof)(x) \) and \( (fog)(x) \).
We have \( (gof)(x) = 3\cos^2 x \) and \( (fog)(x) = \cos(3x^2) \).
Clearly, the expressions are different. For example, let's test for \( x = 0 \):
\( (gof)(0) = 3\cos^2(0) = 3(1)^2 = 3 \).
\( (fog)(0) = \cos(3 \cdot 0^2) = \cos(0) = 1 \).
Since \( 3 \neq 1 \), we have shown that \( (gof)(x) \neq (fog)(x) \).
Step 4: Final Answer:
Since \( 3\cos^2 x \neq \cos(3x^2) \) for all \( x \in \mathbb{R} \), it is proven that \( gof \neq fog \).