Composition of Functions: Additional Practice (Set 2)

Here $(f \circ g)(x) = \sqrt{x^2} = \lvert x \rvert$; find where it is defined.

$x^2 \ge 0 \text{ for every real } x$

The inner output $x^2$ is always nonnegative, so $g(x)$ is always in $\operatorname{dom} f$, and the domain is all real numbers. Contrast $\operatorname{dom}(g \circ f) = \{x \ge 0\}$.

Apply $f$ twice.

$f^2(x) = f(3 - x) = 3 - (3 - x) = x$

So $f \circ f$ returns $x$: applying this $f$ twice undoes it, though $f$ itself is not the identity.

The output map is everything applied after $f$: multiply by $2$, then add $1$.

$A(y) = 2y + 1$

So $y = 2f(3x - 6) + 1 = (A \circ f \circ B)(x)$ with $A(y) = 2y + 1$.

Compose inside out: $f$, then $g$, then $h$.

$h(g(f(x))) = h((x - 1)^2) = 2(x - 1)^2 = 2x^2 - 4x + 2$

Associativity lets us write $h \circ g \circ f$ with no parentheses.

Translations always commute; a scaling with a shift need not.

$(f \circ g)(x) = 2(x + 3) = 2x + 6, \qquad (g \circ f)(x) = 2x + 3$

These differ, so that pair fails to commute; the other three are pairs of translations, which always commute.

The output map is applied after $f$: here just subtract $5$.

$y = f(2x) - 5 = (A \circ f \circ B)(x), \quad A(y) = y - 5,\ B(x) = 2x$

The factor $2$ is inside, so it belongs to $B$, not $A$.

The identity returns its input unchanged, so it leaves $f$ alone on either side.

$f \circ \operatorname{id} = f, \qquad \operatorname{id} \circ f = f$

The one caution is that $\operatorname{id}$ must be taken on the correct set so the domains match, since equality of functions needs equal domains.

Need $x \ne 0$ for $f$, and $f(x) = \tfrac{1}{x} \ne 1$ so $g$ accepts it.

$\frac{1}{x} = 1 \iff x = 1, \quad\text{so exclude } x = 0 \text{ and } x = 1$

Both $x = 0$ (undefined $f$) and $x = 1$ (where $f(x) = 1$, barred by $g$) are removed.

Need $x - 2 \ge 0$ for $f$, and $f(x) = \sqrt{x - 2} \ne 0$ for $g(x) = \tfrac{1}{x}$.

$\sqrt{x - 2} \ne 0 \iff x \ne 2, \quad\text{with}\quad x - 2 \ge 0 \implies x > 2$

The root must be defined and nonzero, so $x > 2$; the value $x = 2$ makes the denominator zero.

Composition is not always commutative, but not never either.

$g \circ f \ne f \circ g \text{ in general}, \qquad g \circ f = f \circ g \text{ for special pairs}$

Translations commute, most pairs do not, so equality holds only for special pairs.

The composite $\dfrac{1}{\sqrt{x} - 2}$ is undefined where the denominator is zero.

$\sqrt{x} - 2 = 0 \iff \sqrt{x} = 2 \iff x = 4$

So $x = 4$ is excluded, even though $4 \ge 0$; the input $x = 2$ is fine since $\sqrt{2} \ne 2$.

The input map $B$ acts before $f$ and the output map $A$ acts after, so $B$ is rightmost.

$y = a\,f(b(x - h)) + k = (A \circ f \circ B)(x)$

with $B(x) = b(x - h)$ and $A(y) = ay + k$; reading right to left, $B$ first, then $f$, then $A$.