Composition of Functions: Practice

Substitute $f(x) = 2x + 1$ into $g(x) = x^2$ and expand.

$(g \circ f)(x) = (2x + 1)^2 = 4x^2 + 4x + 1$

The reverse order gives $(f \circ g)(x) = 2x^2 + 1$.

In $f \circ g$ the inner function is $g(x) = x^2$, so substitute into $f(x) = 2x + 1$.

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

The reverse order gives $(g \circ f)(x) = (2x + 1)^2 = 4x^2 + 4x + 1$.

In $f \circ g$, $g$ acts first, so put $g(x) = x + 3$ into $f(x) = x^2$.

$(f \circ g)(x) = (x + 3)^2$

Expanding gives $x^2 + 6x + 9$, not $x^2 + 9$; the reverse order gives $(g \circ f)(x) = x^2 + 3$.

Need $x \in \operatorname{dom} f$, so $x \ge 0$, and $f(x) \in \operatorname{dom} g$, so $\sqrt{x} \ne 5$.

$\sqrt{x} = 5 \iff x = 25$

So exclude $x = 25$. The trap $x \ne 5$ is the intersection of domains; $x = 5$ is fine because $\sqrt{5} \ne 5$.

We need some function $f$ with $g(f(x)) = (f(x))^4$ equal to $(3x - 2)^4$, and $f(x) = 3x - 2$ is one such function.

$f(x) = 3x - 2 \ \Rightarrow\ (f(x))^4 = (3x - 2)^4$

It is the only one of the four choices that works; the fourth power forces just $\lvert f(x) \rvert = \lvert 3x - 2 \rvert$, so the natural inner base is $3x - 2$.

$f^2 = f \circ f$, so substitute $f(x) = x^2 - 1$ into $f$.

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

The output square $(f(x))^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1$ is a different function.

A decomposition needs $g(f(x)) = \tfrac{1}{x + 3}$.

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

The others give $\tfrac{1}{x} + 3$, $\tfrac{1}{x - 3}$, and $x + 2$.

$T \circ S$ means do $S$ first (stretch), then $T$ (shift).

$(T \circ S)(y) = T(3y) = 3y + 2$

The reverse order $(S \circ T)(y) = 3(y + 2) = 3y + 6$, so the two orders differ.

With outer $g(x) = \sqrt{x}$, the inner $f$ must supply what is under the root.

$g(f(x)) = \sqrt{x - 1} \iff f(x) = x - 1$

Peeling off the square root leaves $x - 1$ as the inside step.

Read the composition from the inside: first $f(2)$, then $g$ of that.

$f(2) = 5, \qquad g(5) = 7$

So $(g \circ f)(2) = g(f(2)) = 7$.

$f^2 = f \circ f$, so apply $f$ twice.

$f^2(x) = f(x - 2) = (x - 2) - 2 = x - 4$

Not the output square $(x - 2)^2$; the superscript counts compositions.

Adding $5$ after $f$ is an output map applied to $y = f(x)$.

$y = f(x) + 5 = (A \circ f)(x), \quad A(y) = y + 5$

The output map $A(y) = y + 5$ acts on the value $f$ returns.