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Middle Years

4.05 Inverse functions (Enrichment)

Lesson

Inverse functions

We call $+7$+7 and $-7$7 inverse operations, because they do the opposite of each other. 

Similarly, functions that do the opposite of each other are called inverse functions. For example the function $f(x)=2x$f(x)=2x and $g(x)=\frac{x}{2}$g(x)=x2 are inverse functions because if we multiply a value by $2$2, then divide it by $2$2 we return to the original value. Another example of a pair of inverse functions is:

$f(x)=x^3$f(x)=x3 and $g(x)=\sqrt[3]{x}$g(x)=3x because finding the cube root is the opposite function to cubing a number.

For a function $f$f, the notation $f^{-1}$f1 is used for the inverse function.

Exploration

If $f(x)=2x$f(x)=2x and $f^{-1}(x)=\frac{x}{2}$f1(x)=x2 determine the following:

 $f(-2),f(0),f(3),f(10)$f(2),f(0),f(3),f(10) and  $f^{-1}(-4),f^{-1}(0),f^{-1}(6),f^{-1}(20)$f1(4),f1(0),f1(6),f1(20)

Do: $f(-2)=-4,f(0)=0,f(3)=6,f(10)=20$f(2)=4,f(0)=0,f(3)=6,f(10)=20

       $f^{-1}(-4)=-2,f^{-1}(0)=0,f^{-1}(6)=3,f^{-1}(20)=10$f1(4)=2,f1(0)=0,f1(6)=3,f1(20)=10

Think: If we were to write these as coordinate pairs we would get the following:

 $f(x)$f(x): $(-2,4),(0,0),(3,6),(10,20)$(2,4),(0,0),(3,6),(10,20)

 $f^{-1}(x)$f1(x): $(4,-2),(0,0),(6,3),(20,10)$(4,2),(0,0),(6,3),(20,10)

Reflect: Look at the order of the $x$x and $y$y-values. Can you see that they are swapped for $f(x)$f(x) and $f^{-1}(x)$f1(x)? This is because $f(x)$f(x) and $f^{-1}(x)$f1(x) are inverse functions. Inverse functions have opposite coordinate pairs

Therefore finding the inverse of a function involves swapping the order of the $x$x and $y$y-values in the ordered pairs. This can be done algebraically by swapping $x$x and $y$y values, or graphically by a reflection over the line $y=x$y=x. This graphical method works because a reflection over the line $y=x$y=x swaps all $x$x and $y$y values, except those sitting on the "mirror line" of $y=x$y=x.

 

Worked example

EXAMPLE 1

Consider the set of coordinates$(-1,-5),(0,-3),(1,-1),(2,1)$(1,5),(0,3),(1,1),(2,1). Graph this set, find the inverse relation and sketch the graph of the inverse on the same axes.

Think: This question requires swapping $x$x and $y$y values.

Do: Swap the coordinates in each pair to give the new set: $(-5,-1),(-3,0),(-1,1),(1,2)$(5,1),(3,0),(1,1),(1,2). This is the set of points of the inverse relation, which is graphed below:

The dotted line $y=x$y=x is included here to illustrate the reflection of the original relation (red) and the inverse (blue).

Find the inverse function algebraically

The steps to finding an inverse function are as follows:

  1. Write $f(x)$f(x) as $y$y.
  2. Swap $x$x with $y$y and $y$y with $x$x.
  3. Make $y$y the subject in the equation of the function.
  4. Replace $y$y with $f^{-1}(x)$f1(x) notation.
  5. Check that the inverse is in fact a function using the vertical line test.

 

Worked examples

EXAMPLE 2

Find the inverse for the function $f(x)=3x+1$f(x)=3x+1. Graph $f$f and its inverse on the same set of axes.

Think: Let $f(x)=y$f(x)=y. This question requires swapping $x$x and $y$y and solving the resulting equation for $y$y.

Do: Swapping $x$x and $y$y gives $x=3y+1$x=3y+1. Solving for $y$y leads to the inverse $y=\frac{x-1}{3}$y=x13. The original function $y$y and its inverse are plotted below:

The intersection between $y$y and its inverse is found by setting $3x+1=\frac{x-1}{3}$3x+1=x13 and solving for $x=\frac{-1}{2}$x=12. Since the intersection will always sit on the line $y=x$y=x (why?), the coordinate is $(\frac{-1}{2},\frac{-1}{2})$(12,12). Note that the inverse relation passes the vertical line test, which means that the function $f(x)=3x+1$f(x)=3x+1 has an inverse function $f^{-1}(x)=\frac{x-1}{3}$f1(x)=x13

 

EXAMPLE 3

Find the inverse for the function $y=x^2+4$y=x2+4. Graph $y$y and its inverse on the same set of axes and state whether the inverse is a function or relation.

Think: This question again requires swapping $x$x and $y$y and solving the equation for $y$y.

Do: Swapping $x$x and $y$y gives $x=y^2+4$x=y2+4. Solving for $y$y leads to the inverse $y=\pm\sqrt{x-4}$y=±x4. The original function $y$y and its inverse are plotted below:

Note that the inverse here is a relation and not a function as it fails the vertical line test. For the inverse to be a function, the input values (domain) of the original function would need to be restricted, for example to only positive $x$x-values. 

 

Practice questions

QUESTION 1

Examine the two curves, shown in the graph below.

Loading Graph...
A graph of two curves from distinct functions on a Cartesian coordinate system, with both the x-axis and y-axis labeled and ranging from -4 to 4. If both curves pass the vertical line test AND are reflections of each other about the line $y=x$y=x, then they are inverse functions of each other.
  1. Are the curves in the graph inverse functions of each other?

    Yes

    A

    No

    B

QUESTION 2

Consider the function $f\left(x\right)=6x^3$f(x)=6x3.

  1. Find an expression for the inverse function $f^{-1}\left(x\right)$f1(x). You may let $y=f^{-1}\left(x\right)$y=f1(x).

Question 3

Below we have sketched the line $y=\frac{x^2}{4}+1$y=x24+1 as defined for $x\le0$x0 (labelled $B$B) over the line $y=x$y=x (labelled $A$A).

Loading Graph...

  1. By reflecting this arm of $y=\frac{x^2}{4}+1$y=x24+1 about the line $y=x$y=x, graph the inverse of the arm of $y=\frac{x^2}{4}+1$y=x24+1 defined over $x\le0$x0.

    Loading Graph...

Question 4

The function $t=\sqrt{\frac{d}{4.9}}$t=d4.9 can be used to find the number of seconds it takes for an object in Earth's atmosphere to fall $d$d metres.

  1. State the function for $d$d in terms of $t$t.

  2. Use your answer from part (a) to find the distance a skydiver has fallen $5$5 seconds after jumping out of a plane.

Does the inverse function exist?

Remember that we define a function as a rule that assigns each $x$x-value to only one $y$y-value.

There's no guarantee that a function always has an inverse function. For example, a function like $y=x^2$y=x2, takes a value of $x$x, and squares it, so the points $(2,4)$(2,4) and $(-2,4)$(2,4) are points on the original function . However when we swap the coordinates to form the inverse function we get $(4,2)$(4,2) and $(4,-2)$(4,2) so we now have two $y$y-values for the same $x$x-value. Our inverse is NOT a function! 

There are two values of $x$x that correspond to $y=4$y=4.

One-to-one functions

A function must have one $x$x-value for every $y$y-value in order for an inverse function to exist. We call these functions one-to-one or injective.

To tell if a function is one-to-one, we draw a horizontal line. If this line line cuts the function only once for all horizontal lines then it's one-to-one. As in the example above, $y=x^2$y=x2 is not one-to-one since the horizontal line $y=4$y=4 cuts the function twice.

Horizontal line test

Functions must be one-to-one in order to for an inverse to exist.

We can check this by making sure that function is only cut once by any horizontal line.

Worked example

example 4

Which of the following functions have an inverse?

$y=x^3,y=x+5,y=\sin x,y=e^x$y=x3,y=x+5,y=sinx,y=ex

Think: We can sketch each function, or use a graphics calculator or other technology to draw each one and pass an imaginary horizontal line through each one. If the function is only cut once, then it is one-to-one and has an inverse function.

Do: Each of the functions are shown below. We can see they all have inverses except for $y=\sin x$y=sinx, where the function is cut more than once by the horizontal line.

Practice question

qUESTION 5

Consider the function given by $f\left(x\right)=\frac{2x+3}{2}$f(x)=2x+32.

  1. Sketch the graph of $f\left(x\right)$f(x) on the coordinate plane below:

    Loading Graph...

  2. Is the function $f\left(x\right)$f(x) one-to-one?

    No

    A

    Yes

    B
  3. Does an inverse function exist for $f\left(x\right)$f(x)?

    No

    A

    Yes

    B

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