Composition of Functions: Practice

Compute the formula and then the domain.

$(g \circ f)(x) = (\sqrt{x})^2 = x, \qquad \operatorname{dom}(g \circ f) = \{x \ge 0\}$

The formula is $x$, but the domain is only $x \ge 0$, while the identity has domain all reals, so by the equality test they differ. The rule $\lvert x \rvert$ is the other order, $f \circ g$.

Need $x \ne 3$ for $f$, and $f(x) = \tfrac{1}{x - 3} \ge 0$ for $g = \sqrt{\phantom{x}}$.

$\frac{1}{x - 3} \ge 0 \iff x - 3 > 0 \iff x > 3$

A reciprocal is nonnegative only when its denominator is positive, and it can never be zero, so $x > 3$.

Compose each and compare with $h(x) = \sqrt{3x + 1}$.

$g(f(x)) = \sqrt{3x + 1} + 1 \ne \sqrt{3x + 1} \quad\text{for } f(x) = 3x + 1,\ g(x) = \sqrt{x} + 1$

The other three each compose to exactly $\sqrt{3x + 1}$, so they are valid; decomposition is not unique, but this one adds a stray $+1$.

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

$h(g(f(x))) = h(2(x + 1)) = (2x + 2)^2 = 4x^2 + 8x + 4$

Associativity means the grouping does not matter, so $h \circ g \circ f$ is unambiguous.

The input map is the inside, factored as $b(x - h)$.

$3x - 6 = 3(x - 2), \quad\text{so}\quad B(x) = 3(x - 2)$

That is the map applied to $x$ before $f$; the output map is $A(y) = 2y + 1$.

Apply $f$ twice: $f(f(x)) = f\!\left(\tfrac{1}{x}\right)$.

$f\!\left(\frac{1}{x}\right) = \frac{1}{\,1/x\,} = x, \qquad x \ne 0$

The formula is $x$, but the domain keeps $x \ne 0$, so $f \circ f$ is not the identity on all reals.

A valid split needs $g(f(x)) = (x^2 + 1)^3$.

$g(f(x)) = (x^2 + 1)^3 \quad\text{when}\quad f(x) = x^2 + 1,\ g(x) = x^3$

The others give $x^6 + 1$, $x^6 + 1$, and $(x + 1)^6$, none equal to $(x^2 + 1)^3$.

Compose: $g(f(x)) = \dfrac{1}{f(x)}$ with $f(x) = \dfrac{1}{x - 1}$.

$g\!\left(\frac{1}{x - 1}\right) = \frac{1}{\,1/(x - 1)\,} = x - 1, \qquad x \ne 1$

The formula simplifies to $x - 1$, yet $x = 1$ stays excluded because $f$ is undefined there.

Composition acts right to left, and $B$ is the rightmost map.

$(A \circ f \circ B)(x) = A(f(B(x)))$

So $B$ acts on $x$ first, then $f$, then $A$.

Iterate: $f^2(x) = 2(2x - 1) - 1 = 4x - 3$, then apply $f$ once more.

$f^3(x) = 2(4x - 3) - 1 = 8x - 7$

Each step doubles the coefficient, matching $f^n(x) = 2^n x - (2^n - 1)$.

Compute both orders and subtract.

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

The difference is $-2x$, zero only at $x = 0$, so the two compositions are genuinely different functions.

Collect the laws proved in this lesson.

$(h \circ g) \circ f = h \circ (g \circ f), \quad f \circ \operatorname{id} = f, \quad g \circ f \ne f \circ g \text{ in general}$

Composition is associative and has an identity, and it is not always commutative, though special pairs commute. Inverses exist only for some functions, the subject of the next lesson.