1.2 - Reversing a function

Inverse function

For a one–one function $f$, the inverse function $f^{-1}$ maps each output of $f$ back to its unique original input. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

The notation $f^{-1}(x)$ means the inverse function's output, not the reciprocal $1/f(x)$.

Worked example 1: undoing a linear function

Question. Find the inverse of $f(x)=3x-4$, $x\in\mathbb R$.

Plan. Solve $y=3x-4$ for $x$, then exchange input and output roles.

The original linear function is one–one and has range $\mathbb R$.

Answer. $f^{-1}(x)=(x+4)/3$, with domain and range $\mathbb R$.

Check. $f^{-1}(f(x))=(3x-4+4)/3=x$. In the other order, $f(f^{-1}(x))=3(x+4)/3-4=x$.

Worked example 2: swapping domain and range

Question. Find the inverse of $g(x)=3x-4$, $1<x\le5$, stating its domain and range.

Plan. Find the original range, then use the rearranged linear rule.

Since the rule increases with $x$,

The same rearrangement as before gives the original input as $(y+4)/3$.

Answer. $g^{-1}(x)=(x+4)/3$, with domain $-1<x\le11$ and range $1<g^{-1}(x)\le5$.

Check. $g^{-1}(11)=5$, so the included endpoint is preserved. Input $-1$ to the inverse would return $1$, excluded from the original domain, so $-1$ is excluded from the inverse domain.

Inverse identities

For a one–one function $f$,

Not in the formula booklet: learn it.

Watch out

If $f(x)=3x-4$, the reciprocal is $1/(3x-4)$, while the inverse is $(x+4)/3$. Check an inverse by composition: it must undo the original function.

Worked example 3: the right-hand branch

Question. Find the inverse of $f(x)=(x+1)^2-2$, $x\ge-1$, stating its domain.

Plan. Rearrange the square and use $x+1\ge0$ to choose its sign.

The original minimum $-2$ is attained at $x=-1$, so its range is $y\ge-2$.

Answer. $f^{-1}(x)=-1+\sqrt{x+2}$, for $x\ge-2$.

Check. $f(1)=2$ and $f^{-1}(2)=-1+2=1$. The inverse always returns a number at least $-1$, as required.

Worked example 4: the left-hand branch

Question. Find the inverse of $g(x)=(x+1)^2-2$, $x\le-1$, stating its domain.

Plan. Use the original condition $x+1\le0$ after rearranging.

The original range is again $y\ge-2$, since the vertex is included.

Answer. $g^{-1}(x)=-1-\sqrt{x+2}$, for $x\ge-2$.

Check. $g(-3)=2$, and $g^{-1}(2)=-1-2=-3$. Choosing the plus sign would return $1$, outside the original domain.

Worked example 5: undoing a restricted cube

Question. Find the inverse of $h(x)=(x+1)^3-2$, $x<-1$, stating its domain.

Plan. Take the real cube root, and find the range from the negative value of $x+1$.

For $x<-1$, $x+1<0$, so $(x+1)^3<0$ and $y<-2$. Every output below $-2$ occurs.

Answer. $h^{-1}(x)=\sqrt[3]{x+2}-1$, for $x<-2$.

Check. $h(-3)=-10$, and $h^{-1}(-10)=\sqrt[3]{-8}-1=-3$. Input $-2$ would return the excluded endpoint $-1$.

Worked example 6: a reciprocal square

Question. Find the inverse of $k(x)=1/(x+1)^2-2$, $x<-1$, stating its domain.

Plan. Isolate $(x+1)^2$, then choose the negative root because $x+1<0$.

The original range is $y>-2$: the reciprocal square is positive, can be arbitrarily small or large, and is never zero.

Answer. $k^{-1}(x)=-1-1/\sqrt{x+2}$, for $x>-2$.

Check. $k(-2)=-1$, and $k^{-1}(-1)=-2$. More generally, if $x>-2$, the inverse returns a number below $-1$, and substituting it into $k$ gives $x+2-2=x$.

Watch out

The domain $x<-1$ in the reciprocal example describes original inputs. It selects the minus sign in the inverse rule. It is not the inverse domain: that comes from the original range $y>-2$.

Graph of an inverse

For a one–one function, reflect its graph in the mirror line $y=x$ to obtain the graph of its inverse. Included and excluded endpoints retain their inclusion status as their coordinates swap.

Worked example 7: reflecting a straight line

Question. Describe a sketch of $f(x)=2x+1$, $x\in\mathbb R$, and its inverse, identifying both graphs' intercepts and their mirror line.

Plan. Find the original intercepts, swap their coordinates, and use the inverse rule to check them.

The original graph passes through $(-1/2,0)$ and $(0,1)$. Reflection gives $(0,-1/2)$ and $(1,0)$.

Rearranging $y=2x+1$ gives $f^{-1}(x)=(x-1)/2$.

Answer. Draw the original line with gradient $2$, its inverse with gradient $1/2$, and the mirror line $y=x$. Original intercepts are $(-1/2,0),(0,1)$; inverse intercepts are $(1,0),(0,-1/2)$.

Check. The inverse rule gives $f^{-1}(0)=-1/2$ and $f^{-1}(1)=0$, confirming the reflected intercepts.

Worked example 8: reflecting asymptotes

Question. For $k(x)=1/(x+1)^2-2$, $x<-1$, describe the graph of its inverse, including its domain, range, asymptotes and the reflection of $(-2,-1)$.

Plan. Swap the original domain and range, then reflect each feature in $y=x$.

The original curve has vertical asymptote $x=-1$, horizontal asymptote $y=-2$, domain $x<-1$ and range $y>-2$.

Answer. The inverse has domain $x>-2$, range $y<-1$, vertical asymptote $x=-2$ and horizontal asymptote $y=-1$. The point $(-2,-1)$ becomes $(-1,-2)$. Draw the reflected branch and the mirror line $y=x$.

Check. Its rule $k^{-1}(x)=-1-1/\sqrt{x+2}$ gives $k^{-1}(-1)=-2$. As $x$ approaches $-2$ from above, the denominator approaches zero and the output decreases without bound; as $x$ increases, the output approaches $-1$ from below.

Exam tip

An inverse sketch should show the correct branch, its key features and the mirror line $y=x$. Reflect the stated function's domain as well as its visible shape.