Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
Inverse functions, domain and range
Reverse a one-to-one function and state where the inverse is defined.
Notes
Before you start: rearrange a linear equation and distinguish an input from an output. A domain is the set of permitted inputs; a range is the set of outputs actually attained.
An inverse reverses the input-output mapping. It is a function only when each attained output comes from one input: the original function must be one-to-one on its stated domain. In \(f^{-1}(x)\), the exponent \(-1\) denotes an inverse, not a reciprocal.
Write \(y=f(x)\), solve for \(x\), then exchange the variable names. The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\). Check both the algebra and these sets.
For a fractional function, retain the excluded denominator value. Rewriting \((ax+b)/(x-c)\) as a constant plus a nonzero multiple of \(1/(x-c)\) can reveal a missing range value.
Recall later
Find the inverse of \(7-2x\) for \(0\le x\le3\). Check: \((7-x)/2\), with inverse domain \([1,7]\).
Common mistakes
Reading inverse notation as 1/f(x).
Giving the old domain to the inverse without finding the original range.
Cancelling or rearranging away an excluded denominator value.
Three worked examples
Example 1
Find the inverse of \(f(x)=4x-7\), for real \(x\).
Show the reasoning
From \(y=4x-7\), \(x=(y+7)/4\). Therefore \(f^{-1}(x)=(x+7)/4\), with domain and range both \(\mathbb R\). Substitution gives \(f(f^{-1}(x))=x\).
Example 2
Find the inverse of \(f(x)=(x+1)/(x-2)\), \(x\ne2\).
Show the reasoning
\(y(x-2)=x+1\), so \(x(y-1)=2y+1\). Thus \(f^{-1}(x)=(2x+1)/(x-1)\), with domain \(x\ne1\). The original form \(1+3/(x-2)\) shows why its range excludes 1.
Example 3
Find the range and inverse domain of \(f(x)=9-2x\), \(1\le x\le5\).
Show the reasoning
The function decreases. Its endpoint outputs are 7 and −1, so its range is \([-1,7]\). The inverse is \((9-x)/2\), defined for \(-1\le x\le7\), with range \([1,5]\).
Apply functions in the correct order and check the intermediate value.
Notes
Before you start: substitute an expression into a formula and solve a linear inequality.
We use \(fg(x)=f(g(x))\): apply \(g\) first, then \(f\). It is not multiplication. Replace every input in the outer formula by the entire inner expression, using brackets.
The domain of \(fg\) consists of inputs in the domain of \(g\) whose outputs lie in the domain of \(f\). For example, a square-root outer function requires the complete inner expression to be nonnegative.
In general \(fg\ne gf\). When solving \(fg(x)=gf(x)\), use the common domain. Inverse compositions give the identity only on the appropriate original or inverse domain.
Recall later
If \(f(x)=x+4\) and \(g(x)=2x\), compare \(fg(x)=2x+4\) and \(gf(x)=2x+8\).
Common mistakes
Treating fg as a product or reversing the agreed order.
Substituting without brackets before squaring.
Checking only the final simplified formula and overlooking the intermediate domain.
Three worked examples
Example 1
Let \(f(x)=3x+2\), \(g(x)=x^2-1\). Find \(fg\) and \(gf\).
Show the reasoning
\(fg(x)=3(x^2-1)+2=3x^2-1\). In the other order, \(gf(x)=(3x+2)^2-1=9x^2+12x+3\). Both domains are \(\mathbb R\).
Example 2
Let \(f(x)=\sqrt{x}\), \(g(x)=x-4\). Find both composite domains.
Show the reasoning
\(fg(x)=\sqrt{x-4}\) requires \(x\ge4\). But \(gf(x)=\sqrt{x}-4\) only requires \(x\ge0\). The same two functions can produce different domains in different orders.
Example 3
For \(f(x)=2x+3\), verify \(ff^{-1}(x)=x\).
Show the reasoning
The inverse is \(f^{-1}(x)=(x-3)/2\). Hence \(f(f^{-1}(x))=2[(x-3)/2]+3=x\). All real inputs are permitted here.
Choose a one-to-one branch and use the correct square-root sign.
Notes
Before you start: complete the square and interpret a vertex.
A full quadratic parabola is not one-to-one: inputs on opposite sides of its vertex can have the same output. Restricting the domain to one side of the vertex gives a one-to-one function.
For \(f(x)=(x-h)^2+k\) on \(x\ge h\), solving \(y=f(x)\) gives \(x=h+\sqrt{y-k}\). On \(x\le h\), use \(x=h-\sqrt{y-k}\). A function inverse uses one branch, not both signs.
The inverse domain is \(x\ge k\) for either full half-parabola. Its range remembers the chosen original branch. Additional endpoint restrictions must also be transferred.
Recall later
For \(f(x)=(x-5)^2+1\), \(x\le5\), the inverse is \(5-\sqrt{x-1}\), defined for \(x\ge1\).
Common mistakes
Keeping both square-root signs in a function inverse.
Choosing the positive branch despite an original input below the vertex.
Giving the inverse range as all real numbers after restricting the original domain.
Three worked examples
Example 1
Find the inverse of \(f(x)=(x+1)^2\), \(x\ge-1\).
Show the reasoning
Since the original input satisfies \(x+1\ge0\), take the positive root: \(x=\sqrt y-1\). Thus \(f^{-1}(x)=\sqrt{x}-1\), with domain \(x\ge0\) and range \(y\ge-1\).
Example 2
Find the inverse of \(f(x)=(x-4)^2+2\), \(x\le4\).
Show the reasoning
Here \(x-4\le0\), so \(x=4-\sqrt{y-2}\). The inverse is \(4-\sqrt{x-2}\), with domain \(x\ge2\) and range \(y\le4\).
Example 3
Why does \(f(x)=x^2\) on \(\mathbb R\) have no function inverse?
Show the reasoning
Both −2 and 2 map to 4. Reversing this would give two outputs for one input. On the restricted domain \(x\ge0\), the inverse is \(\sqrt{x}\); on \(x\le0\), it is \(-\sqrt{x}\).
Reflect coordinates and interpret common points without assuming a false general rule.
Notes
Before you start: plot coordinates and solve simultaneous equations.
If \((a,b)\) lies on \(y=f(x)\), then \((b,a)\) lies on \(y=f^{-1}(x)\). The graphs are reflections in \(y=x\). Swap vertices and endpoints as well as ordinary points; transfer domain and range restrictions.
A fixed point solves \(f(x)=x\) and lies on both graphs. For a strictly increasing real function, every common point must be on \(y=x\): if \(f(x)=y\) and \(f(y)=x\), then \(x
Do not use that conclusion for every invertible function. A decreasing function can have common points off \(y=x\); for example \(f(x)=6-x\) is its own inverse, so the two entire graphs coincide.
Recall later
Reflect \((-2,5)\) in \(y=x\): the image is \((5,-2)\). The signs stay attached to their coordinates.
Common mistakes
Reflecting in an axis rather than y=x.
Swapping the coordinates but dropping a branch restriction.
Assuming every intersection with an inverse is a fixed point, even for decreasing functions.
Three worked examples
Example 1
The graph of \(f(x)=2x+2\) contains \((1,4)\) and \((3,8)\). Locate the reflected points.
Show the reasoning
The inverse graph contains \((4,1)\) and \((8,3)\). Its equation is \(y=(x-2)/2\). The common fixed point solves \(2x+2=x\), giving \((-2,-2)\).
Example 2
Reflect the branch \(y=(x-2)^2+1\), \(x\ge2\).
Show the reasoning
Its vertex \((2,1)\) becomes \((1,2)\); its point \((3,2)\) becomes \((2,3)\). The inverse is \(y=2+\sqrt{x-1}\), \(x\ge1\), with \(y\ge2\).
Example 3
Find the inverse of \(f(x)=6-x\) and describe the two graphs.
Show the reasoning
Rearranging \(y=6-x\) gives \(x=6-y\), so \(f^{-1}(x)=6-x\). The graphs coincide. For example \((1,5)\) is common even though it is not on \(y=x\).
Track changes inside and outside a function, including domains and ranges.
Notes
Before you start: read a graph's domain and range and rearrange a linear equation.
Changes outside \(f\) act on outputs: \(af(x)+b\) maps \(y\) to \(ay+b\). Changes inside act on inputs: for \(f(cx+d)\), a point with old input \(u\) has new input \((u-d)/c\), provided \(c\ne0\).
Thus \(f(x-h)\) moves a graph right by \(h\); \(f(cx)\) gives a horizontal scale factor \(1/|c|\), with reflection in the y-axis if \(c<0\). A negative outside multiplier reflects in the x-axis. Describe an ordered sequence or give an unambiguous coordinate rule.
Find the new domain by putting \(cx+d\) in the old domain. Find the new range by applying \(ay+b\) to the old range; reverse the endpoint order if \(a<0\).
Recall later
Under \(y=2f(x-1)+3\), the point \((4,-2)\) moves to \((5,-1)\).
Common mistakes
Using the inside scale factor instead of its reciprocal.
Translating vertically before stretching when the given formula requires the other order.
Forgetting to transform domain or range endpoints.
Three worked examples
Example 1
Describe \(y=f(x+2)-3\) using a point \((u,v)\).
Show the reasoning
The new coordinates are \((u-2,v-3)\): translate 2 units left and 3 units down. The inside plus sign moves the graph left.
Example 2
Find the image of \((6,-1)\) under \(y=2f(3x)\).
Show the reasoning
Solve \(3x=6\), so the new input is 2. The new output is \(2(-1)=-2\). The image is \((2,-2)\): horizontal scale factor 1/3, vertical scale factor 2.
Example 3
Find the vertex and range of \(y=-(x+1)^2+4\).
Show the reasoning
Starting with \(x^2\), translate left 1, reflect in the x-axis, then translate up 4. The vertex is \((-1,4)\). The range is \(y\le4\).