Inverse Functions (Edexcel IGCSE Maths A (Modular))

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Mark Curtis

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Inverse Functions

What is an inverse function?

An inverse function does the opposite (reverse) operation of the function it came from
- E.g. If a function “doubles the number then adds 1”
- Then its inverse function “subtracts 1, then halves the result”
  - The same inverse operations are used when solving an equation or rearranging a formula
An inverse function performs the inverse operations in the reverse order

What notation is used for inverse functions?

The inverse function of $f (x)$ is written as $f^{- 1} (x) = \dots$ or $f^{- 1} : x \mapsto \dots$
- For example, if $f (x) = 2 x + 1$
- The inverse function is $f^{- 1} (x) = \frac{x - 1}{2}$ or $f^{- 1} : x \mapsto \frac{x - 1}{2}$
If $f (a) = b$ then $f^{- 1} (b) = a$
- For example
  - $f (3) = 2 \times 3 + 1 = 7$ (inputting 3 into $f$ gives 7)
  - $f^{- 1} (7) = \frac{7 - 1}{2} = 3$ (inputting 7 into $f^{- 1}$ gives back 3)

How do I find an inverse function algebraically?

The process for finding an inverse function is as follows:
- Write the function as $y = . . .$
  - E.g. The function $f (x) = 2 x + 1$ becomes $y = 2 x + 1$
- Swap the $x$ s and $y$ s to get $x = \dots$
  - E.g. $x = 2 y + 1$
  - The letters change but no terms move
- Rearrange the expression to make $y$ the subject again
  - E.g. $x = 2 y + 1$ becomes $x - 1 = 2 y$ so $y = \frac{x - 1}{2}$
- Replace $y$ with $f^{- 1} (x) = \dots$ (or $f^{- 1} : x \mapsto \dots$ )
  - E.g. $f^{- 1} (x) = \frac{x - 1}{2}$
  - This is the inverse function
  - $y$ should not appear in the final answer

The composite function of $f$ followed by $f^{- 1}$ (or the other way round) cancels out
- ${ff}^{- 1} (x) = f^{- 1} f (x) = x$
  - If you apply a function to x, then apply its inverse function, you get back x
  - Whatever happened to x gets undone
  - f and f^-1 cancel each other out when applied together
For example, solve $f^{- 1} (x) = 5$ where $f (x) = 2^{x}$
- Finding the inverse function $f^{- 1} (x)$ algebraically in this case is tricky
  - (It is impossible if you haven't studied logarithms!)
- Instead, you can take $f$ of both sides of $f^{- 1} (x) = 5$ and use the fact that ${ff}^{- 1}$ cancel each other out:
  - $\begin{array}{rcl} {ff}^{- 1} (x) & = & f (5) \end{array}$ which cancels to $x = f (5)$ giving $x = 2^{5} = 32$

How do I find the domain and range of an inverse function?

The domain of an inverse function has exactly the same values as the range of the original function
- E.g. If $f (x) = \frac{3}{x + 1}$ has a range of $f (x) > 5$
  - then its inverse function, $f^{- 1} (x) = \frac{3}{x} - 1$ , has the domain $x > 5$
  - Remember to always write domains in terms of $x$
The range of an inverse function has exactly the same values as the domain of the original function
- E.g. If $f (x) = \frac{3}{x + 1}$ has a domain of $x < - 1$
  - then its inverse function, $f^{- 1} (x) = \frac{3}{x} - 1$ , has the range $f^{- 1} (x) < - 1$
  - Remember to always write ranges in terms of their function, $f^{- 1} (x)$

Worked Example

A function $f (x) = 5 - 3 x$ has the domain $- 2 < x \leq 7$ .

(a) Use algebra to find $f^{- 1} (x)$ .

Write the function in the form $y = 5 - 3 x$ and then swap the $x$ and $y$

$y = 5 - 3 x x = 5 - 3 y$

Rearrange the expression to make $y$ the subject again

$\begin{array}{rcl} x & = & 5 - 3 y \\ x + 3 y & = & 5 \\ 3 y & = & 5 - x \\ y & = & \frac{5 - x}{3} \end{array}$

Rewrite the answer using inverse function notation

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

(b) Find the domain of $f^{- 1} (x)$ .

The domain of the inverse function is the range of the original function

Find the range of $f (x)$ by first finding $f (- 2)$ and $f (7)$

$\begin{array}{rcl} f (- 2) & = & 5 - 3 (- 2) = 5 + 6 = 11 \\ f (7) & = & 5 - 3 (7) = 5 - 21 = - 16 \end{array}$

The graph of $y = 5 - 3 x$ is a straight line with a negative gradient
Between x = -2 and x = 7 the graph decreases from a height of 11 to a height of -16

The range of $f (x)$ is $- 16 \leq f (x) < 11$

Note that the inequality is "equal to" at x = 7, f(x) = -16
(this is the opposite order of "equal to" in the domain)

The domain of $f^{- 1} (x)$ takes the same values as range of $f (x)$
Write down the domain of $f^{- 1} (x)$
(Remember that domains are always written in terms of $x$ )

$- 16 \leq x < 11$

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