Q.If $f(x) = x^2 - 4$ and $g(x) = x+3$, find $(f \circ g)(x)$ and $(g \circ f)(x)$, and show that $(f \circ g)(x) \neq (g \circ f)(x)$.

Substituting one function into the other, in each of the two possible orders, gives two different resulting formulas.

Finding $(f \circ g)(x)$: here $g$ is applied first, then $f$.

Expanding $(x+3)^2 = x^2+6x+9$:

So $(f\circ g)(x) = x^2+6x+5$.

Finding $(g \circ f)(x)$: here $f$ is applied first, then $g$.

So $(g\circ f)(x) = x^2-1$.

Showing they are not equal: compare the two formulas directly:

These are different polynomials (they differ by the term $6x+6$ at every $x$). As a concrete check, at $x=1$: $(f\circ g)(1) = 1+6+5=12$, while $(g\circ f)(1) = 1-1=0$ — two clearly different values. …

A common algebra slip is expanding $(x+3)^2$ incorrectly as $x^2+9$ (forgetting the middle term $6x$) — always expand a squared binomial as …