Graphs of Inverse Functions: Free Response

Swapping coordinates reflects each point across $y=x$, so $(a,b)$ becomes $(b,a)$ for every point on $h$.

$(-3,-1)\to(-1,-3), \quad (1,1)\to(1,1), \quad (3,4)\to(4,3), \quad (5,6)\to(6,5).$

The second point does not move, because its two coordinates were already equal.

Reflecting across $y=x$ trades the two axes, so the domain and range of $h$ swap roles for $h^{-1}$.

$\text{domain of } h^{-1} = \text{range of } h = [-1,6], \qquad \text{range of } h^{-1} = \text{domain of } h = [-3,5].$

Among the four points, $(1,1)$ has equal coordinates, so it already sits on the mirror line $y=x$; a point on the mirror is its own reflection, and swapping two equal numbers leaves them unchanged.

The reflection across $y=x$ is a swap of the plane's two axes: the horizontal axis (all points $(t,0)$) lands on the vertical axis (all points $(0,t)$), and the reverse.

$\text{domain of } f = \text{horizontal spread}, \qquad \text{range of } f = \text{vertical spread}.$

After the reflection, whatever used to be a horizontal spread becomes a vertical spread and whatever used to be vertical becomes horizontal. Since domain is defined as the horizontal spread and range as the vertical one, the reflected graph's domain must be the original's vertical spread (its range), and its range must be the original's horizontal spread (its domain). No other outcome is possible once the axes themselves are what moved.

$k^{-1}(-1)$ asks which input of $k$ produced the output $-1$; the point $(0,-1)$ answers it directly.

$(0,-1) \text{ on } k \ \Longrightarrow\ (-1,0) \text{ on } k^{-1}, \text{ so } k^{-1}(-1) = 0.$

Similarly $(4,7)$ on $k$ swaps to $(7,4)$ on $k^{-1}$, so $k^{-1}(7) = 4$.

$k^{-1}(x) = 6$ says $6$ is the output of $k^{-1}$ at input $x$, which by the swap means $(x,6)$ is on $k^{-1}$ and so $(6,x)$ is on $k$.

$(6,11) \text{ on } k \ \Longrightarrow\ (11,6) \text{ on } k^{-1}, \text{ so } k^{-1}(11) = 6.$

The input that makes $k^{-1}(x)=6$ true is $x=11$.

The defining property of an inverse is that $b=k(a)$ and $a=k^{-1}(b)$ say the identical thing about the same pair of numbers, just naming the input and output the other way around.

$(a,b)\text{ on }k \iff b=k(a) \iff a=k^{-1}(b) \iff (b,a)\text{ on }k^{-1}.$

So a point of $k$ is not evidence FOR a fact about $k^{-1}$; it already IS a fact about $k^{-1}$, once its coordinates are read in the swapped order. No formula for either function ever enters the chain.

Apply $p(x)=x^2$ at each restricted input.

$p(0)=0^2=0, \quad p(-1)=(-1)^2=1, \quad p(-2)=(-2)^2=4.$

So the three points are $(0,0)$, $(-1,1)$, $(-2,4)$, all with $x \le 0$ as required.

Swap the coordinates of each point from part A.

$(0,0)\to(0,0), \quad (-1,1)\to(1,-1), \quad (-2,4)\to(4,-2).$

The restricted $p$ has domain $x\le 0$ and range $y\ge 0$ (every output of a square is nonnegative). Reflecting swaps those, so the inverse has domain $x\ge 0$ (the old range) and range $y\le 0$ (the old domain).

The claim to test is whether the curve from part B could equal $y=\sqrt{x}$, whose range is $y\ge 0$.

$\text{range of the inverse in part B} = \text{domain of the restricted } p = \{x \le 0\}.$

That range is never positive, while $\sqrt{x}$'s range is never negative; the two ranges only share the single value $0$. Since the two functions' outputs disagree almost everywhere, they are not the same function. The reason traces directly to which branch was reflected: the RIGHT branch ($x\ge 0$) is what the lesson reflects to reach $y=\sqrt{x}$, and this question reflected the LEFT branch instead, which lands the mirror image in the opposite half of the plane.

Swap $x$ and $y$ in $y = 6-x$ and solve for $y$.

$x = 6 - y \ \Longrightarrow\ y = 6 - x.$

So $f^{-1}(x) = 6-x$, identical to $f(x)$. Every rule of the form $f(x) = c - x$ is its own inverse.

Pick any input other than the fixed point and check both graphs.

$f(1) = 6-1 = 5,$

so $(1,5)$ lies on the graph of $f$. Because part A showed $f^{-1}(x) = 6-x = f(x)$, the graph of $f^{-1}$ is the identical curve, so $(1,5)$ lies on the graph of $f^{-1}$ too, with no reflecting required: it is already on both. Since its coordinates $1$ and $5$ are not equal, it is not on $y=x$. So the two graphs share this point, and it is not on the diagonal, which is exactly what the claim in the stem says cannot happen.

The weaker, true claim is that wherever $f$ crosses $y=x$, that point is shared with $f^{-1}$ too. Find where $f$ crosses the diagonal.

$6-x = x \ \Longrightarrow\ 6 = 2x \ \Longrightarrow\ x = 3,$

so $(3,3)$ is the crossing point, and it is on both graphs, consistent with the weaker claim. Part B's point $(1,5)$ is a separate shared point, off the diagonal. Both facts hold at once: the guaranteed diagonal point is still there, and the claim that ONLY that kind of point can be shared is the part that fails. Refuting only never requires denying always.

Both labeled points on the line $y=1$ satisfy $w(x)=1$:

$w(-2)=1, \qquad w(2)=1.$

So the single horizontal line $y=1$ meets the graph of $w$ at TWO different points, $(-2,1)$ and $(2,1)$. That is exactly what it means to fail the horizontal line test: some horizontal line crosses the graph more than once, so two different inputs, $-2$ and $2$, share the same output.

Swap the coordinates of each of the three points on the restricted branch.

$(0,-1)\to(-1,0), \quad (2,1)\to(1,2), \quad (3,4)\to(4,3).$

On $x\ge 0$, $w$ has domain $0\le x\le 3$ and range $-1\le y\le 4$ (reading from the endpoints $(0,-1)$ and $(3,4)$). Reflecting swaps those, so the inverse has domain $-1\le x\le 4$ and range $0\le y\le 3$.

Part A found that the failure was caused by ONE pair of points, $(-2,1)$ and $(2,1)$, sharing an output. Restricting to $x\ge 0$ keeps only $(2,1)$ from that pair and discards $(-2,1)$, so no two remaining inputs share an output on that branch; it is one-to-one, and a one-to-one graph's reflection always passes the vertical line test, by the same argument the lesson gives for the horizontal-to-vertical swap.

Restricting to $x\le 0$ instead keeps $(-2,1)$ and discards $(2,1)$, which repairs the SAME conflict from the other side; that branch is equally one-to-one for the same reason, so it also gives a valid inverse graph. But its points are different, for instance $(-2,1)$ reflects to $(1,-2)$, a point nowhere on part B's inverse:

$(-2,1)\to(1,-2) \quad \text{vs.} \quad (2,1)\to(1,2).$

Two valid restrictions, both one-to-one, reflect to two different graphs, because they are reflections of different branches.