iGCSE (2021 Edition)

Lesson

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`)=2`x` and $g(x)=\frac{x}{2}$`g`(`x`)=`x`2 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`)=`x`3 and $g(x)=\sqrt[3]{x}$`g`(`x`)=^{3}√`x` because finding the cube root is the opposite function to cubing a number.

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

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

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

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

$f^{-1}(x)$`f`−1(`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 $f^{-1}(x)$`f`−1(`x`)? This is because $f(x)$`f`(`x`) and $f^{-1}(x)$`f`−1(`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`.

As the $x$`x` and $y$`y`-values are opposites, this means the domain and range of inverse functions are also opposites.

Domain and range for $f(x)$`f`(`x`) and $f^{-1}(x)$`f`−1(`x`)

The domain of a function is the range of the inverse function.

The range of a function is the domain of the inverse function.

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, stating its domain and range.

**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).

For this blue inverse set, we can see that the domain = $\left\{-5,-3,-1,1\right\}${−5,−3,−1,1} and the range = $\left\{-1,0,1,2\right\}${−1,0,1,2}, which is the opposite of the original red set.

The steps to finding an inverse function are as follows:

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

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

**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`=3`y`+1. Solving for $y$`y` leads to the inverse $y=\frac{x-1}{3}$`y`=`x`−13. 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}$3`x`+1=`x`−13 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`)=3`x`+1 has an inverse function $f^{-1}(x)=\frac{x-1}{3}$`f`−1(`x`)=`x`−13.

The domain and range for both $f(x)$`f`(`x`) and its inverse are all real numbers in $x$`x` and $y$`y`.

Find the inverse for the function $y=x^2+4$`y`=`x`2+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`=`y`2+4. Solving for $y$`y` leads to the inverse $y=\pm\sqrt{x-4}$`y`=±√`x`−4. 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 domain of the original function would need to be restricted, for example to only positive $x$`x`-values.

Examine the two curves, shown in the graph below.

Are the curves in the graph inverse functions of each other?

Yes

ANo

B

Consider the function given by $f\left(x\right)=x+6$`f`(`x`)=`x`+6 defined over the interval $\left[0,\infty\right)$[0,∞).

Plot the function $f\left(x\right)=x+6$

`f`(`x`)=`x`+6 over its domain.Loading Graph...Find the inverse $f^{-1}$

`f`−1.State the domain and range of $f^{-1}$

`f`−1 in interval notation.Domain: $\editable{}$

Range: $\editable{}$

Plot the function $f^{-1}$

`f`−1 over its domain.Loading Graph...

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

Find an expression for the inverse function $f^{-1}\left(x\right)$

`f`−1(`x`). You may let $y=f^{-1}\left(x\right)$`y`=`f`−1(`x`).

Below we have sketched the line $y=\frac{x^2}{4}+1$`y`=`x`24+1 as defined for $x\le0$`x`≤0 (labelled $B$`B`) over the line $y=x$`y`=`x` (labelled $A$`A`).

Loading Graph...

By reflecting this arm of $y=\frac{x^2}{4}+1$

`y`=`x`24+1 about the line $y=x$`y`=`x`, graph the inverse of the arm of $y=\frac{x^2}{4}+1$`y`=`x`24+1 defined over $x\le0$`x`≤0.Loading Graph...

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

State the function for $d$

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

Remember that we define a function as a rule that assigns each $x$`x`-value in our domain 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`=`x`2, 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!

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`=`x`2 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.

Which of the following functions have an inverse?

$y=x^3,y=x+5,y=\sin x,y=e^x$`y`=`x`3,`y`=`x`+5,`y`=`s``i``n``x`,`y`=`e``x`

**Think:** We can sketch each function, or use our graphics calculator 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`=`s``i``n``x`, where the function is cut more than once by the horizontal line.

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

Sketch the graph of $f\left(x\right)$

`f`(`x`) on the coordinate plane below:Loading Graph...Is the function $f\left(x\right)$

`f`(`x`) one-to-one?No

AYes

BDoes an inverse function exist for $f\left(x\right)$

`f`(`x`)?No

AYes

B

Understand the terms: function, domain, range (image set), one-one function, inverse function and composition of functions.

Use the notation f ^(–1)(x).

Use the notation f^2(x) [= f(f(x))].

Explain in words why a given function is a function or why it does not have an inverse.

Find the inverse of a one-one function.

Form composite functions.

Sketch graphs to show the relationship between a function and its inverse.