Inverse Functions: Practice

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

$2y = x + 1 \;\Rightarrow\; x = 2y - 1$

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

$f^{-1}(12)$ is the input $x$ with $f(x) = 12$.

$5x + 2 = 12 \;\Rightarrow\; x = 2$

So $f^{-1}(12) = 2$; the value $62$ is $f(12)$, the wrong direction.

Inverses undo each other, so both compositions must return $x$.

$g(f(x)) = (x + 7) - 7 = x, \qquad f(g(x)) = (x - 7) + 7 = x$

The pair $x^2$ and $\sqrt{x}$ fails on all reals since $\sqrt{x^2} = |x|$, and $3 - x$ gives $g(f(x)) = 3 - (x - 3) = 6 - x \ne x$.

A rule is one-to-one only if no two inputs share an output, so exhibit a collision for each non-answer.

$(-1)^2 - 1 = 1^2 - 1 = 0, \qquad |1 - 2| = |3 - 2| = 1$

So $x^2 - 1$ repeats at $\pm 1$ and $|x - 2|$ repeats at $1$ and $3$, while the constant $f(x) = 5$ sends every input to $5$. Only the line $2x + 1$ never repeats an output, so only it is one-to-one on all reals.

An inverse must undo $f$ from both sides, so both compositions must equal the identity.

$g \circ f = \operatorname{id} \text{ and } f \circ g = \operatorname{id}$

Checking only one side can pass while the other fails, as with $x^2$ and $\sqrt{x}$.

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

$xy = 2 \;\Rightarrow\; x = \frac{2}{y}$

So $f^{-1}(x) = \frac{2}{x}$: this function is its own inverse. The choice $\frac{x}{2}$ is the inverse of $2x$, a different function.

The inverse sends the output back to the input.

$f(2) = 9 \;\Rightarrow\; g(9) = 2$

So $g(9) = 2$.

Solve $y = -x + 6$ for $x$.

$x = 6 - y = -y + 6$

So $f^{-1}(x) = -x + 6$: the function is its own inverse. Indeed $f(f(x)) = 6 - (6 - x) = x$.

The domain of the inverse is the range of the original.

$\operatorname{dom} f^{-1} = \operatorname{range} f = [2, \infty)$

Domain and range trade places under inversion.

Solve $y = \frac{x - 4}{3}$ for $x$.

$3y = x - 4 \;\Rightarrow\; x = 3y + 4$

So $f^{-1}(x) = 3x + 4$.

A constant sends every input to the same output, so it is far from one-to-one.

$f(0) = f(1) = 7, \text{ but } 0 \ne 1$

With outputs colliding, no rule can recover the input, so $f$ has no inverse.

Solve $y = 2x^3 + 1$ for $x$, undoing the operations in reverse.

$y - 1 = 2x^3 \;\Rightarrow\; x^3 = \frac{y - 1}{2} \;\Rightarrow\; x = \sqrt[3]{\frac{y - 1}{2}}$

So $f^{-1}(x) = \sqrt[3]{\frac{x - 1}{2}}$.