Inverse Functions (Cambridge (CIE) IGCSE Maths)

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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 straight f open parentheses x close parentheses is written as space straight f to the power of negative 1 end exponent left parenthesis x right parenthesis equals horizontal ellipsis space space

    • For example, if straight f left parenthesis x right parenthesis equals 2 x plus 1

    • The inverse function is straight f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator 2 end fraction  or straight f to the power of negative 1 end exponent colon space x rightwards arrow from bar fraction numerator x minus 1 over denominator 2 end fraction

  • If straight f open parentheses a close parentheses equals b then straight f to the power of negative 1 end exponent open parentheses b close parentheses equals a

    • For example

      • straight f open parentheses 3 close parentheses equals 2 cross times 3 plus 1 equals 7 (inputting 3 into straight f gives 7)

      • straight f to the power of negative 1 end exponent open parentheses 7 close parentheses equals fraction numerator 7 minus 1 over denominator 2 end fraction equals 3 (inputting 7 into straight f to the power of negative 1 end exponent 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 bold italic y bold equals bold. bold. bold.

      • E.g. The function straight f left parenthesis x right parenthesis equals 2 x plus 1 becomes y equals 2 x plus 1

    • Swap the xs and ys to get x equals horizontal ellipsis

      • E.g. x equals 2 y plus 1

      • The letters change but no terms move

    • Rearrange the expression to make bold italic y the subject again

      • E.g. x equals 2 y plus 1 becomes x minus 1 equals 2 y so y equals fraction numerator x minus 1 over denominator 2 end fraction

    • Replace bold italic y with space straight f to the power of negative 1 end exponent left parenthesis x right parenthesis equals horizontal ellipsis space space(or straight f to the power of negative 1 end exponent colon space x rightwards arrow from bar horizontal ellipsis)

      • E.g. straight f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 1 over denominator 2 end fraction

      • This is the inverse function

      • y should not appear in the final answer

  • The composite function of straight f followed by straight f to the power of negative 1 end exponent (or the other way round) cancels out

    • ff to the power of negative 1 end exponent open parentheses x close parentheses equals straight f to the power of negative 1 end exponent straight f open parentheses x close parentheses equals 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 straight f to the power of negative 1 end exponent open parentheses x close parentheses equals 5 where straight f open parentheses x close parentheses equals 2 to the power of x

    • Finding the inverse function straight f to the power of negative 1 end exponent open parentheses x close parentheses algebraically in this case is tricky

      • (It is impossible if you haven't studied logarithms!)

    • Instead, you can take straight f of both sides of straight f to the power of negative 1 end exponent open parentheses x close parentheses equals 5 and use the fact that ff to the power of negative 1 end exponent cancel each other out:

      • table row cell ff to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell straight f open parentheses 5 close parentheses end cell end table which cancels to x equals straight f open parentheses 5 close parentheses giving x equals 2 to the power of 5 equals 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 straight f open parentheses x close parentheses equals fraction numerator 3 over denominator x plus 1 end fraction has a range of straight f open parentheses x close parentheses greater than 5

      • then its inverse function, straight f to the power of negative 1 end exponent open parentheses x close parentheses equals 3 over x minus 1, has the domain x greater than 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 straight f open parentheses x close parentheses equals fraction numerator 3 over denominator x plus 1 end fraction has a domain of x less than negative 1

      • then its inverse function, straight f to the power of negative 1 end exponent open parentheses x close parentheses equals 3 over x minus 1, has the range straight f to the power of negative 1 end exponent open parentheses x close parentheses less than negative 1

      • Remember to always write ranges in terms of their function, straight f to the power of negative 1 end exponent open parentheses x close parentheses

Worked Example

A function straight f open parentheses x close parentheses equals 5 minus 3 xhas the domain negative 2 less than x less or equal than 7.

(a) Use algebra to find straight f to the power of negative 1 end exponent open parentheses x close parentheses.

Write the function in the form y equals 5 minus 3 x and then swap the x and y

y equals 5 minus 3 x
x equals 5 minus 3 y

Rearrange the expression to make y the subject again

table row x equals cell 5 minus 3 y end cell row cell space x plus 3 y end cell equals 5 row cell 3 y end cell equals cell 5 minus x end cell row y equals cell fraction numerator 5 minus x over denominator 3 end fraction end cell end table

Rewrite the answer using inverse function notation

Error converting from MathML to accessible text.

(b) Find the domain of straight f to the power of negative 1 end exponent open parentheses x close parentheses.

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

Find the range of straight f open parentheses x close parentheses by first finding straight f open parentheses negative 2 close parentheses and straight f open parentheses 7 close parentheses

table row cell straight f open parentheses negative 2 close parentheses end cell equals cell 5 minus 3 open parentheses negative 2 close parentheses equals 5 plus 6 equals 11 end cell row cell straight f open parentheses 7 close parentheses end cell equals cell 5 minus 3 open parentheses 7 close parentheses equals 5 minus 21 equals negative 16 end cell end table

The graph of y equals 5 minus 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 straight f open parentheses x close parentheses is negative 16 less or equal than straight f open parentheses x close parentheses less than 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 straight f to the power of negative 1 end exponent open parentheses x close parentheses takes the same values as range of straight f open parentheses x close parentheses
Write down the domain of straight f to the power of negative 1 end exponent open parentheses x close parentheses
(Remember that domains are always written in terms of x)

bold minus bold 16 bold less or equal than bold italic x bold less than bold 11

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Naomi C

Author: Naomi C

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Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

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Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.