Q.Find $gof$ and $fog$, if $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are given by $f(x) = \cos x$ and $g(x) = 3x^2$. Show that $gof \neq fog$.

Concept: Domain of composite function — both $f$ and $g$ map $\mathbb{R} \to \mathbb{R}$, so all compositions are defined on $\mathbb{R}$.

Step 1: Compute $gof$.

$gof(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3\cos^2 x$.

Step 2: Compute $fog$.

$fog(x) = f(g(x)) = f(3x^2) = \cos(3x^2)$.

Step 3: Compare. …

The key idea here is that composition of functions is not commutative — the order in which you apply the functions matters. When we write $gof$, we read it as "g after f": first do $f$, then feed that result into $g$. For $fog$, it's the reverse: first $g$, then $f$.

Let's build each composition step by step.

1. Finding $gof$ We start with $gof(x) = g(f(x))$. Since $f(x) = \cos x$, we put this into $g$:

So $gof$ squares the cosine and multiplies by 3.

2. Finding $fog$ Now $fog(x) = f(g(x))$. Since $g(x) = 3x^2$, we put this into $f$:

So $fog$ takes the cosine of $3x^2$.

3. Comparing the two At a glance, $3\cos^2 x$ and $\cos(3x^2)$ look very different. But to be rigorous, we show they differ at a specific input. Take $x = 0$: $gof(0) = 3\cos^2 0 = 3 \cdot 1^2 = 3,$ …

Method: Finding $g\circ f$, $f\circ g$ and showing they differ

Use this when asked to compute both composites of two real functions and prove composition is not commutative.

Steps

Step 1: Build $g\circ f$

$(g\circ f)(x)=g(f(x))$: substitute the whole expression for $f(x)$ into $g$'s rule and simplify.

Step 2: Build $f\circ g$

$(f\circ g)(x)=f(g(x))$: substitute $g(x)$ into $f$'s rule. Keep the order straight — the function on the right acts first.

Step 3: Show inequality with a single test point …

Common Mistakes

Mistake 1: Confusing composition with multiplication or reversing the order

Why it's wrong: $g\circ f$ is $g(f(x))=3\cos^2 x$, not $\cos x\cdot 3x^2$ and not $\cos(3x^2)$. Correct approach: substitute the inner function into the outer one, keeping $f$ (right side) acting first.

Mistake 2: Claiming $g\circ f\ne f\circ g$ "because they look different" without proof …