IGCSEMathsEdexcelMaths A (Modular)Higher Unit 2Revision NotesFunctions, Graphs & DifferentiationFunctionsInverse Functions

Inverse Functions (Edexcel IGCSE Maths A (Modular))

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Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Edexcel Maths A (Modular): Higher Unit 2Cambridge (CIE) International Maths: Extended

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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 or  straight f to the power of negative 1 end exponent colon space x rightwards arrow from bar horizontal ellipsis

    • 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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    Author
    Reviewer
    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    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.