Explore: Inverses and a Composition of Functions Solutions

Inverses and a Composition of Functions Solutions

A composition of functions can be used to determine algebraically if two functions are inverses .
If $[f \circ g] (x) = x$ and $[g \circ f] (x) = x$ , then the functions are inverses.
However, if both compositions do not equal x , then the functions are not inverses.
Both $[f \circ g] (x)$ and $[g \circ f] (x)$ must be checked because the composition of functions is not commutative .

Example 6

Determine which of the given functions are inverses.

$p (x) = \frac{\sqrt{x + 1}}{2} q (x) = 4 (x^{2} - 1), x \geq 0 r (x) = 4 x^{2} - 1, x \geq 0$

$\begin{array}{rcl} [p \circ q] (x) & = & p [q (x)] = \frac{\sqrt{4 (x^{2} - 1) + 1}}{2}, x \geq 0 \\ = & \frac{\sqrt{4 x^{2} - 4 + 1}}{2} \\ = & \frac{\sqrt{4 x^{2} - 3}}{2} \end{array}$

Not inverses

$\begin{array}{rcl} [p \circ r] (x) & = & p [r (x)] = \frac{\sqrt{4 x^{2} - 1 + 1}}{2}, x \geq 0 \\ = & \frac{\sqrt{4 x^{2}}}{2} \\ = & \frac{2 x}{2} \\ = & x \end{array}$

$\begin{array}{rcl} r [p (x)] & = & 4 {(\frac{\sqrt{x + 1}}{2})}^{2} - 1 \\ = & 4 (\frac{x + 1}{4}) - 1 \\ = & x + 1 - 1 \\ = & x \end{array}$

Because $p [r (x)] = x$ AND $r [p (x)] = x$ , the functions p(x) and r(x) are inverses.

Note

Because there are domain restrictions for given functions, there may also be domain restrictions for the composition.