Inverse Functions: Additional Practice (Set 2)

If the codomain is larger than the range, a one-to-one function can still miss some declared outputs, so no inverse exists there.

$f \circ g = \operatorname{id} \text{ fails at any } b \text{ with } b \notin \operatorname{range} f$

Taking the codomain to be the range makes 'onto' automatic, leaving one-to-one as the exact condition.

On $x \ge 2$ the base $x - 2$ is nonnegative, so the square root recovers it.

$y = (x - 2)^2 \;\Rightarrow\; \sqrt{y} = x - 2 \;\Rightarrow\; x = \sqrt{y} + 2$

So $f^{-1}(x) = \sqrt{x} + 2$.

Solve $y = \frac{x + 2}{x - 1}$ for $x$.

$y(x - 1) = x + 2 \;\Rightarrow\; x(y - 1) = y + 2 \;\Rightarrow\; x = \frac{y + 2}{y - 1}$

So $f^{-1}(x) = \frac{x + 2}{x - 1} = f(x)$: this function is its own inverse.

A fixed point $f(a) = a$ gives $(a, a)$ on $f$, and $f^{-1}(a) = a$ puts it on $f^{-1}$ too.

$f(2) = 2 \;\Rightarrow\; (2, 2) \text{ lies on both graphs}$

The other points are not fixed points, so nothing forces them onto both graphs.

The notation $(f(x))^{-1}$ means the reciprocal of the value, not the inverse function.

$(f(x))^{-1} = \frac{1}{\,1/x\,} = x$

Here the compositional inverse $f^{-1}(x) = \frac{1}{x}$ happens to look the same as $f$, but $(f(x))^{-1} = x$ is a separate computation.

Setting $f(x) = f^{-1}(x)$ leads to $x^9 = x$.

$x^9 - x = x(x^8 - 1) = 0 \;\Rightarrow\; x \in \{-1, 0, 1\}$

These give $(0, 0)$ on $y = x$, plus $(1, -1)$ and $(-1, 1)$ off it, for three shared points.

Inverting swaps the two sets a function lives between.

$\operatorname{dom} f^{-1} = \operatorname{range} f, \qquad \operatorname{range} f^{-1} = \operatorname{dom} f$

So the range of $f^{-1}$ is the domain of $f$. Equal graphs happen only for self-inverse functions, not in general.

Because $f$ is increasing, any crossing with $f^{-1}$ lies on $y = x$, so solve $f(x) = x$.

$2x - 8 = x \;\Rightarrow\; x = 8$

So the graphs meet at $(8, 8)$, which is on $y = x$.

Chain the two inverse properties through associativity.

$g = g \circ (f \circ h) = (g \circ f) \circ h = h$

So any two inverses coincide, which is why $f^{-1}$ is well defined.

A quadratic turns around, so it repeats outputs.

$x^2 - 4x = (x - 2)^2 - 4, \quad \text{so } f(0) = f(4) = 0$

With two inputs sharing an output, $x^2 - 4x$ is not one-to-one on $\mathbb{R}$; the line and the two increasing cubics are.

Start from two points in the range of $f$, which is the domain of $f^{-1}$. Take $u < v$ in the range, and write $u = f(a)$ and $v = f(b)$, so $a = f^{-1}(u)$ and $b = f^{-1}(v)$. Suppose $a \ge b$; because $f$ is increasing, that forces $f(a) \ge f(b)$.

$a \ge b \;\Rightarrow\; u = f(a) \ge f(b) = v$

That contradicts $u < v$, so instead $a < b$, which says $f^{-1}(u) < f^{-1}(v)$. Hence the inverse of an increasing one-to-one function is increasing.

Solve $y = \frac{5}{x} - 1$ for $x$.

$y + 1 = \frac{5}{x} \;\Rightarrow\; x = \frac{5}{y + 1}$

So $f^{-1}(x) = \frac{5}{x + 1}$.