Inverse Functions: Practice

Swap $x$ and $y$, then clear the fraction and solve.

$x = \frac{y}{2} - 4 \;\Longrightarrow\; x + 4 = \frac{y}{2} \;\Longrightarrow\; y = 2x + 8$

The domain of $f^{-1}$ equals the range of $f$, because the outputs of $f$ become the inputs of $f^{-1}$.

$\text{domain}(f^{-1}) = \text{range}(f) = \text{all real numbers except } 0$

Swap $x$ and $y$, then solve for $y$ by taking the reciprocal.

$x = \frac{1}{y} \;\Longrightarrow\; y = \frac{1}{x}$

So $\dfrac{1}{x}$ is its own inverse.

Swap $x$ and $y$, then solve for $y$.

$x = 6y - 12 \;\Longrightarrow\; x + 12 = 6y \;\Longrightarrow\; y = \frac{x + 12}{6}$

Finding $f^{-1}(9)$ means solving $f(x) = 9$ for the input.

$2x + 1 = 9 \;\Longrightarrow\; 2x = 8 \;\Longrightarrow\; x = 4$

Swap $x$ and $y$, take reciprocals, then isolate $y$.

$x = \frac{2}{y + 1} \;\Longrightarrow\; y + 1 = \frac{2}{x} \;\Longrightarrow\; y = \frac{2}{x} - 1 = \frac{2 - x}{x}$

The domain of the inverse is the range of $f$, which is the nonnegative numbers.

$f^{-1}(x) = x^2, \quad x \ge 0$

Without the restriction, $x^2$ would not undo the square root correctly.

Swap $x$ and $y$, then solve for $y$.

$x = -y + 3 \;\Longrightarrow\; y = 3 - x$

The inverse is $3 - x$, which is the same rule again.

The inverse is $f^{-1}(x) = 2x + 4$, so evaluate it at $3$.

$f^{-1}(3) = 2(3) + 4 = 10$

Check: $f(10) = \dfrac{10 - 4}{2} = 3$.

Swap $x$ and $y$, then solve for $y$.

$x = 8 - 2y \;\Longrightarrow\; 2y = 8 - x \;\Longrightarrow\; y = \frac{8 - x}{2}$

Because the inverse swaps inputs and outputs, the outputs of $f$ (its range) become the inputs of $f^{-1}$ (its domain).

$\text{domain}(f^{-1}) = \text{range}(f)$

Swap $x$ and $y$, then solve for $y$.

$x = 5y \;\Longrightarrow\; y = \frac{x}{5}$

The reciprocal $\dfrac{1}{5x}$ is not the inverse.