Menu
Sign up Sign in

AI HL · 12.5

Functions

0 of 45 skills done

0
12.5.4

Inverse functions

12.5.4

Inverse functions

Target: I can say that a function and its inverse are reflections in the line y = x, and that their coordinates swap. (+5 more)

By the end of this section

Key points

What an inverse does

  • The inverse f⁻¹ reverses f: it takes an output of f back to the input.
  • The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. Every point (a, b) on f gives the point (b, a) on f⁻¹.
  • ln x and eˣ are inverses of each other, so their graphs are reflections in y = x.

Finding and checking an inverse (HL)

  • To find f⁻¹(x): write y = f(x), make x the subject, then write x in place of y.
  • f(f⁻¹(x)) = f⁻¹(f(x)) = x. Use this to check your answer.
  • f⁻¹(x) is not 1/f(x). The −1 is not a power.

One-to-one functions (HL)

  • A function has an inverse only if it is one-to-one: each output comes from one input.
  • Restrict the domain to make a function one-to-one. Use the calculator to find the turning point first.
  • Choose the side of the turning point the question asks for, such as x ≥ k.

Inverse of a non-linear function (HL)

  • The domain of \(f^{-1}\) is the range of f, and the range of \(f^{-1}\) is the domain of f.
  • For \(f(x)=ae^x+b\), the inverse is \(\ln\!\left(\frac{x-b}{a}\right)\). The horizontal asymptote \(y=b\) of f becomes the vertical asymptote \(x=b\) of \(f^{-1}\).
  • For a quadratic with domain \(x\geq k\), complete the square and take the positive square root. For \(x\leq k\) take the negative root.
  • \(f(x)=f^{-1}(x)\) happens where the graphs meet on the line \(y=x\). Solve \(f(x)=x\) with the calculator.

On the calculator

Finding a minimum or maximum

  • Graph the function in a window that shows the turning point.
  • Use the calculator's minimum (or maximum) tool on the graph.
  • Write down the x-coordinate. That is the k in x ≥ k.

Screenshot and video to come

Solving f(x) = x

  • Enter f(x) as Y1 and x as Y2.
  • Use the intersection tool once for each point where the graphs meet.
  • The x-coordinates solve \(f(x)=f^{-1}(x)\).

Screenshot and video to come

Explore: a function and its inverse

Drag the orange point P along the graph of f. Its mirror image P′ lies on the graph of f⁻¹. Look at the coordinates of both before you read the answer.

f(x)
f⁻¹(x)
Domain of f⁻¹

Drag P, or select it (Tab) and use the arrow keys.

Drag P along the graph of f.

  1. Move P. What do you notice about the coordinates of P and P′? Where would P′ be if P were on the line y = x?
  2. Look at the dashed line from P to P′. What is its angle to the line y = x, and where does it cross that line?
  3. Choose eˣ. What is f⁻¹(1)? Which point of f tells you?
  4. Choose x² and tick the box. Move P to the left of the y-axis. What goes wrong with the reflection, and how does x ≥ 0 fix it?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Finding an inverse

\(f(x)=\frac{5 x + 2}{3}\). Find \(f^{-1}(x)\), then find \(f^{-1}(4)\).

Working
  1. Let y = f(x)

    \(y=\frac{5 x + 2}{3}\)

    Write y for the output.

  2. Make x the subject

    \(3y=5x+2 \;\Rightarrow\; 5x=3y-2 \;\Rightarrow\; x=\frac{3 y - 2}{5}\)

    Undo the operations in reverse order.

  3. Swap y for x

    \(f^{-1}(x)=\frac{3 x - 2}{5}\)

    The input of the inverse is called x.

  4. Evaluate

    \(f^{-1}(4)=2\)

    Check: f(2) = 4, so the inverse takes 4 back to 2.

Video walkthrough coming
Checking an inverse

\(f(x)=\frac{2 x - 1}{3}\). A student says that \(f^{-1}(x)=\frac{3 x - 1}{2}\). Work out \(f(f^{-1}(6))\) and use it to say whether the student is right.

Working
  1. Put the inverse into f

    \(f(f^{-1}(x))=\frac{2\left(\frac{3 x - 1}{2}\right)-1}{3}\)

    For a correct inverse this must equal x.

  2. Simplify

    \(f(f^{-1}(x))=\frac{3 x - 2}{3}\)

    This is not x, so the student is wrong.

  3. At x = 6

    \(f(f^{-1}(6))=\frac{16}{3}=5.33\)

    A correct inverse would give 6 here.

  4. The correct inverse

    \(f^{-1}(x)=\frac{3 x + 1}{2}\)

    Make x the subject of y = f(x). The student had the sign of 1 wrong.

Video walkthrough coming
Restricting the domain

Use your calculator to find the largest domain of the form \(x \geq k\) for which \(f(x)=x^2\ln x\) has an inverse.

0.511.522.5−1−0.50.511.522.53xyy = x² ln x
Working
  1. Why restrict

    A function has an inverse only if it is one-to-one.

    To the left of the minimum the graph goes down, to the right it goes up. A horizontal line can cut it twice.

  2. 0.511.522.5−1−0.50.511.522.53xyy = x² ln x
    Minimum

    From the calculator: minimum at (0.607, −0.184)

    Graph f and find the minimum point.

  3. Keep the right-hand side

    \(x \geq 0.607\)

    To the right of the minimum f is increasing, so it is one-to-one.

Video walkthrough coming
A function and its inverse

The point \((2, e^2)\) lies on the graph of \(y=e^x\). Give the matching point on the graph of \(y=\ln x\).

−22468−22468xyy = eˣy = ln x
Working
  1. −22468−22468xyy = eˣy = ln x
    The mirror line

    y = ln x is the reflection of y = eˣ in the line y = x.

    ln x is the inverse of eˣ.

  2. −22468−22468xyy = eˣy = ln x
    Swap the coordinates

    \((2, e^2) \to (e^2, 2) = (7.39, 2)\)

    A reflection in y = x swaps x and y.

  3. Check

    \(\ln(7.39)=2\)

    The point sits on y = ln x, as it should.

Video walkthrough coming
Inverse of an exponential function

\(f(x)=2e^x+5\). Find \(f^{-1}(x)\) and state its domain and range. Then find \(f^{-1}(11)\) to 3 significant figures.

Working
  1. Let y = f(x)

    \(y=2e^x+5\)

    Write y for the output.

  2. Make x the subject

    \(e^x=\dfrac{y-5}{2} \;\Rightarrow\; x=\ln\!\left(\dfrac{y-5}{2}\right)\)

    Undo the 5, then the 2, then take the logarithm to undo the exponential.

  3. Swap y for x

    \(f^{-1}(x)=\ln\!\left(\dfrac{x-5}{2}\right)\)

    The input of the inverse is called x.

  4. Domain and range

    Range of f: \(y>5\). So the domain of \(f^{-1}\) is \(x>5\) and its range is all real numbers.

    The domain of the inverse is the range of f. The logarithm needs a positive input: \(x-5>0\).

  5. Evaluate

    \(f^{-1}(11)=\ln 3=1.10\)

    Check: \(f(1.0986)=2(3)+5=11\).

Video walkthrough coming
Inverse of a restricted quadratic

\(f(x)=x^2-6x+11\), for \(x\geq3\). Find \(f^{-1}(x)\) and state its domain and range. Then find \(f^{-1}(11)\).

Working
  1. Complete the square

    \(f(x)=(x-3)^2+2\)

    The vertex is at \((3,\,2)\). This is also the left end of the domain.

  2. Make x the subject

    \(y=(x-3)^2+2 \;\Rightarrow\; x-3=\pm\sqrt{y-2}\)

    Two square roots; choose the one that matches the domain.

  3. Choose the root

    \(x\geq3\), so \(x-3\geq0\): \(x=3+\sqrt{y-2}\)

    Take the positive root. Then \(f^{-1}(x)=3+\sqrt{x-2}\).

  4. Domain and range

    Range of f: \(y\geq2\). Domain of \(f^{-1}\): \(x\geq2\). Range of \(f^{-1}\): \(y\geq3\).

    The domain and range swap places.

  5. Evaluate

    \(f^{-1}(11)=3+\sqrt{9}=6\)

    Check: \(f(6)=36-36+11=11\).

Video walkthrough coming
When a function equals its inverse

\(f(x)=\ln x+2\), for \(x>0\). Solve \(f(x)=f^{-1}(x)\). Give the larger solution to 3 significant figures.

−1123456−1123456xyy = f(x)y = f⁻¹(x)
Working
  1. Where they meet

    The graphs of f and \(f^{-1}\) are reflections in \(y=x\). Where they cross, they meet on the line \(y=x\).

    So \(f(x)=f^{-1}(x)\) is the same as \(f(x)=x\).

  2. Solve with the calculator

    \(\ln x+2=x\). Enter \(Y1=\ln x+2\) and \(Y2=x\) and find the intersections.

    You do not need the formula for \(f^{-1}\).

  3. Read both

    From the calculator: \(x=0.159\) and \(x=3.15\)

    Check the graph: both points lie on \(y=x\).

  4. State

    The larger solution is \(x=3.15\)

    Give both solutions if the question asks for all of them.

Video walkthrough coming