Inverse Functions: Free Response

Write the rule as an equation in $y$:

$y = 3x - 5.$

Swap $x$ and $y$, so the letter that was the output now plays the input:

$x = 3y - 5.$

Solve for $y$ by adding $5$ and dividing by $3$:

$x + 5 = 3y \quad\Longrightarrow\quad y = \frac{x+5}{3}.$

So $f^{-1}(x) = \dfrac{x+5}{3}$.

Evaluate each quantity on its own. For $f^{-1}(4)$, substitute $4$ into the formula from part A:

$f^{-1}(4) = \frac{4+5}{3} = \frac{9}{3} = 3.$

For $\dfrac{1}{f(4)}$, first find $f(4)$, then take its reciprocal:

$f(4) = 3(4) - 5 = 7, \qquad \frac{1}{f(4)} = \frac{1}{7}.$

The two results, $3$ and $\dfrac{1}{7}$, are not equal.

Compare what each symbol is built to reverse. $f^{-1}$ is defined by the equation

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

which says $f^{-1}$ undoes composition: composing it with $f$ returns the input unchanged, the identity for composition. The reciprocal $\dfrac{1}{f(x)}$ instead undoes multiplication: it is the number that multiplies with $f(x)$ to give $1$, the identity for multiplication,

$f(x) \cdot \frac{1}{f(x)} = 1.$

Composition and multiplication are different operations, with different identities ($x$ for one, $1$ for the other), so the object that undoes one has no reason to match the object that undoes the other. Part B confirmed this on one pair of numbers; this argument is why it holds for $f(x)=3x-5$ at every input, and for any function with a nonzero output.

Take the two directions in turn.

If $x_0$ belongs to $R$, then $x_0$ is a legal input. By the definition of range, $x_0=f(a)$ for some $a$ in $D$, so $(a,x_0)$ is a pair of $f$. Swapping that pair, as the inverse is built to do, gives $(x_0,a)$ as a pair of $f^{-1}$:

$(a,x_0)\ \text{is a pair of}\ f \;\Longrightarrow\; (x_0,a)\ \text{is a pair of}\ f^{-1}.$

A function has a defined output exactly at the inputs that appear as the first entry of one of its pairs, so $x_0$ is a legal input to $f^{-1}$.

If $x_0$ is a legal input, then $x_0$ belongs to $R$. If $x_0$ is a legal input to $f^{-1}$, then $f^{-1}$ has some pair $(x_0,b)$. By the swap that defines $f^{-1}$, that pair came from the pair $(b,x_0)$ of $f$, which means $x_0=f(b)$ for the input $b$ in $D$. An output produced by some input of $f$ is exactly what belongs to $R$, so $x_0$ belongs to $R$.

Both directions hold, so the legal inputs of $f^{-1}$ are exactly $R$.

Notice that swapping every pair of $f^{-1}$ recovers every pair of $f$: swapping $(a,b)$ to $(b,a)$ and then swapping again returns $(a,b)$. So $f$ is built from $f^{-1}$ by exactly the same swap that built $f^{-1}$ from $f$; in symbols, $(f^{-1})^{-1}=f$.

Part A proved a general fact about ANY one-to-one function and its inverse: the legal inputs of the inverse equal the range of the original. Nothing in that argument used any property specific to $f$; it only used the swap. So the same argument applies with $f^{-1}$ playing the role of the original function and $f$ playing the role of its inverse:

$\text{legal inputs of } (f^{-1})^{-1} = \text{range of } f^{-1}.$

But $(f^{-1})^{-1}=f$, whose legal inputs are $D$ by definition. So

$D = \text{range of } f^{-1}.$

State the answer first, from the theorem alone. Since domain of $f^{-1}$ is $R$ and range of $f^{-1}$ is $D$,

$\text{domain of } f^{-1} = \{x\neq 0\}, \qquad \text{range of } f^{-1} = \{x\neq -4\}.$

Now check it with algebra. Write $y=\dfrac{3}{x+4}$, swap the letters, and solve for $y$:

$x=\frac{3}{y+4} \;\Longrightarrow\; x(y+4)=3 \;\Longrightarrow\; y+4=\frac{3}{x} \;\Longrightarrow\; y=\frac{3}{x}-4.$

So $f^{-1}(x)=\dfrac{3}{x}-4$. This formula is undefined only at $x=0$, matching the stated domain, and its output $\dfrac{3}{x}-4$ can equal every real number except $-4$ (since $\dfrac{3}{x}$ is never $0$), matching the stated range.

Substitute the whole rule for $g$ into $f$:

$f(g(x)) = \left(\sqrt{x+6}\right)^2 - 6 = (x+6) - 6 = x.$

Squaring undoes the root here, but only where the root is defined, so this holds for every $x\ge -6$, the domain of $g$.

Substitute the whole rule for $f$ into $g$:

$g(f(x)) = \sqrt{(x^2-6)+6} = \sqrt{x^2}.$

Evaluate at $x=-4$:

$g(f(-4)) = \sqrt{(-4)^2} = \sqrt{16} = 4.$

The result is $4$, not $-4$, so this direction fails at $x=-4$.

An inverse pair is defined by two equations, not one:

$f(g(x))=x \qquad \text{and} \qquad g(f(x))=x.$

The classmate confirmed only the first. Part B computed the second and found $g(f(x))=\sqrt{x^2}$, which does not simplify to $x$: squaring first and then rooting recovers only the SIZE of a number, not its sign, so it returns $x$ when $x\ge 0$ but returns $-x$ when $x<0$, as the $x=-4$ case showed. Checking one direction can therefore pass by coincidence at every point where the two directions happen to agree, while the other direction is where the failure would show up. Only checking both closes off that possibility, which is exactly why the definition asks for both.

Write the rule as an equation in $y$:

$y=\frac{9}{5}c+32.$

Swap $c$ and $y$:

$c=\frac{9}{5}y+32.$

Solve for $y$ by subtracting $32$ and dividing by $\dfrac{9}{5}$:

$c-32=\frac{9}{5}y \quad\Longrightarrow\quad y=\frac{5}{9}(c-32).$

Renaming the input $x$, the formula that recovers Celsius from a Fahrenheit reading is $f^{-1}(x)=\dfrac{5}{9}(x-32)$.

Substitute $68$ into the formula from part A:

$f^{-1}(68) = \frac{5}{9}(68-32) = \frac{5}{9}(36) = 20.$

Check it going forward, back into the original rule:

$f(20) = \frac{9}{5}(20)+32 = 36+32 = 68,$

which matches the forecast, so $20$ degrees Celsius is confirmed as the reading it came from.

Being a function only guarantees ONE output per input; it says nothing about whether two different inputs could share an output, and it is exactly that possibility that would break an inverse. The property that rules it out is one-to-one.

Confirm $f$ has it algebraically. Suppose $f(a)=f(b)$:

$\frac{9}{5}a+32=\frac{9}{5}b+32 \quad\Longrightarrow\quad \frac{9}{5}a=\frac{9}{5}b \quad\Longrightarrow\quad a=b.$

So equal outputs force equal inputs: $f$ is one-to-one, and that is what makes the formula from part A a real inverse rather than a rule that only works on part of the domain.

Pick two inputs the same distance on either side of $x=-4$, since $(x+4)^2$ treats them alike. Try $x=-1$ and $x=-7$:

$f(-1) = (-1+4)^2-9 = 3^2-9 = 0, \qquad f(-7) = (-7+4)^2-9 = (-3)^2-9 = 0.$

Both give $0$. Since two different inputs, $-1$ and $-7$, share the output $0$, $f$ is not one-to-one on all reals. An inverse would have to send $0$ back to a single input, but there are two honest candidates, so no function can do that job, and $f$ has no inverse function on this domain.

Write the restricted rule as an equation in $y$:

$y=(x+4)^2-9, \qquad x\ge -4.$

Swap $x$ and $y$. The restriction was on the INPUT of $f$, which becomes the OUTPUT of $f^{-1}$, so it now attaches to $y$:

$x=(y+4)^2-9, \qquad y\ge -4.$

Solve for $y$:

$x+9=(y+4)^2.$

Two numbers square to give $x+9$, namely $\pm\sqrt{x+9}$, but $y\ge -4$ means $y+4\ge 0$, so only the nonnegative root is allowed:

$y+4=\sqrt{x+9} \quad\Longrightarrow\quad y=\sqrt{x+9}-4.$

So $f^{-1}(x)=\sqrt{x+9}-4$, defined wherever $x+9\ge 0$, that is, $x\ge -9$.

From part B, $f^{-1}(x)=\sqrt{x+9}-4$ has domain $x\ge -9$ (where the root is defined), and since $\sqrt{x+9}\ge 0$, its range is $y\ge -4$.

Now find the range of $f$ on $x\ge -4$. This restricted piece of the parabola opens upward starting at its vertex, so its smallest value is at $x=-4$:

$f(-4) = (-4+4)^2-9 = -9,$

and $f$ only increases for $x>-4$, so the range of the restricted $f$ is $y\ge -9$.

Line the two pairs up. The range of the restricted $f$, $y\ge -9$, matches the domain of $f^{-1}$, $x\ge -9$; and the restricted domain of $f$, $x\ge -4$, matches the range of $f^{-1}$, $y\ge -4$. Both trade places exactly as the general rule predicts.