Graphs of Inverse Functions: Additional Practice (Set 2)

Work from the inside out: $f(3) = 7$, and $f^{-1}$ undoes $f$.

$f^{-1}(f(3)) = f^{-1}(7) = 3$

Running a function and then its inverse returns the original input, no matter the value.

The point $(4, 4)$ is on the mirror line, so it is a fixed point shared by $f$ and $f^{-1}$.

$(4, 4) \to (4, 4) \;\Rightarrow\; f^{-1}(4) = 4$

Where $f$ meets $y = x$, the inverse meets it too.

On $x \le 0$ the outputs are $y \ge 0$, and the inputs are negative. From $y = x^2$ with $x \le 0$, swap to $x = y^2$ with $y \le 0$.

$x = y^2, \; y \le 0 \;\Rightarrow\; y = -\sqrt{x}$

The restriction $y \le 0$ keeps the negative square root, so the reflection is $y = -\sqrt{x}$.

A parabola fails the horizontal line test: two inputs share an output. For $f(x) = x^2 - 2x$, the value at $x = 0$ and $x = 2$ agree.

$f(0) = 0, \qquad f(2) = 4 - 4 = 0$

Since $f(0) = f(2)$, the function is not one-to-one, so its whole reflection is not a function.

For a decreasing function, a point and its swap can both lie on the graph. Suppose $f$ is decreasing and passes through $(0, 2)$ and $(2, 0)$; then $f^{-1}$ passes through the swapped points, which are the same two points.

$(0, 2) \to (2, 0), \qquad (2, 0) \to (0, 2)$

So the graphs share $(0, 2)$ and $(2, 0)$, neither of which is on $y = x$. Increasing functions meet their inverse only on $y = x$.

The solution of $f(x) = x$ marks where $f$ meets the line $y = x$, at $(6, 6)$. A point on the mirror is fixed.

$(6, 6) \to (6, 6)$

So $(6, 6)$ lies on both $f$ and $f^{-1}$.

Reflecting across the $x$-axis negates the output, giving $y = -f(x)$, which does not swap input and output. Reflecting across $y = x$ swaps them.

$(a, b) \xrightarrow{\; y = x \;} (b, a)$

Only the reflection across $y = x$ produces the inverse.

On $x \le 0$ the outputs are $y \ge 0$. From $y = x^2$ with $x \le 0$, swap to $x = y^2$ with $y \le 0$.

$x = y^2, \; y \le 0 \;\Rightarrow\; y = -\sqrt{x}$

The negative branch of the square root is the reflection of the left branch.

The vertex is at $x = 2$, and the parabola is one-to-one only on one side of the vertex.

$\text{restrict to } x \ge 2 \;\Rightarrow\; \text{one-to-one}$

On $x \ge 2$ each height comes from a single input, so the reflection passes the vertical line test.

The horizontal line test checks that $f$ is one-to-one, which is exactly what its reflection needs to pass the vertical line test.

$f \text{ one-to-one} \iff f^{-1} \text{ is a function}$

Without the horizontal line test, the reflection would fail the vertical line test.

The mirror line $y = x$ makes a $45^\circ$ angle, and reflecting across it sends an angle $\theta$ (measured from the $x$-axis) to $90^\circ - \theta$.

$90^\circ - 70^\circ = 20^\circ$

So the inverse's line makes a $20^\circ$ angle with the positive $x$-axis.

Reflection across $y = x$ preserves whether a graph rises or falls, so a decreasing graph reflects to a decreasing graph.

$f \text{ decreasing} \;\Rightarrow\; f^{-1} \text{ decreasing}$

Since the curve is one-to-one, the reflection is a genuine function, and it decreases too.