2 The argument of a function

The input to a function is sometimes called its argument . It is frequently necessary to obtain the output from a function if we are given its argument. For example, given the function g ( t ) = 3 t + 2 we may require the value of the output when the argument is 4. We write this as g ( t = 4 ) or more usually and compactly as g ( 4 ) . In this case the value of g ( 4 ) is 3 × 4 + 2 = 14 .

Example 2

Given the function f ( x ) = 3 x + 1 find

  1. f ( 2 )
  2. f ( − 1 )
  3. f ( 6 )
Solution
  1. The output from the function needs to be found when the argument or input is 2. We need to replace x by 2 in the expression for the function. We find

    f ( 2 ) = 3 × 2 + 1 = 7

  2. Here the argument is − 1 . We find

    f ( − 1 ) = 3 × ( − 1 ) + 1 = − 2

  3. f ( 6 ) = 3 × 6 + 1 = 19 .
Task!

Given the function g ( t ) = 6 t + 4 find

  1. g ( 3 ) ,
  2. g ( 6 ) ,
  3. g ( − 2 )
  1. g ( 3 ) = 6 × 3 + 4 = 22 ,
  2. g ( 6 ) = 40 ,
  3. g ( − 2 ) = − 8

It is possible to obtain the value of a function when the argument is an algebraic expression. Consider the following Example.

Example 3

Given the function y ( x ) = 3 x + 2 find

  1. y ( t )
  2. y ( 2 t )
  3. y ( z + 2 )
  4. y ( 5 x )
Solution

The rule for this function is ‘multiply the input by 3 and then add 2’. We can apply this rule whatever the argument.

  1. In this case the argument is t . Multiplying this by 3 and adding 2 we find y ( t ) = 3 t + 2 . Equivalently we can replace x by t in the expression for the function, so, y ( t ) = 3 t + 2 .
  2. In this case the argument is 2 t . We need to replace x by 2 t in the expression for the function. So y ( 2 t ) = 3 ( 2 t ) + 2 = 6 t + 2
  3. In this case the argument is z + 2 . We find y ( z + 2 ) = 3 ( z + 2 ) + 2 = 3 z + 8 . It is important to note that y ( z + 2 ) is not y × ( z + 2 ) = y z + y 2 but instead reads ‘ y of ( z + 2 ) ’ where ‘of’ means ‘take the function of’.
  4. Here we have a complication. The argument is 5 x and so there appears to be a clash of notation with the original expression for the function. There is no problem if we remember that the rule is to multiply the input by 3 and then add 2. The input now is 5 x . So y ( 5 x ) = 3 ( 5 x ) + 2 = 15 x + 2 .
Task!

Given the function g ( x ) = 8 − 2 x find  

  1. g ( 4 ) ,
  2. g ( 4 t ) ,
  3. g ( 2 x − 3 )
  1. g ( 4 ) = 8 − 2 × 4 = 0
  2. g ( 4 t ) = 8 − 2 × 4 t = 8 − 8 t
  3. g ( 2 x − 3 ) = 8 − 2 ( 2 x − 3 ) = 14 − 4 x
Exercises
  1. Explain what is meant by the ‘argument’ of a function.
  2. Given the function g ( t ) = 8 t + 3 find
    1. g ( 7 ) ,
    2. g ( 2 ) ,
    3. g ( − 0.5 ) ,
    4. g ( − 0.11 )
  3. Given the function f ( t ) = 2 t 2 + 4 find:
    1. f ( x )
    2. f ( 2 x )
    3. f ( − x )
    4. f ( 4 x + 2 )
    5. f ( 3 t + 5 )
    6. f ( λ )
    7. f ( t − λ )
    8. f ( t α )
  4. Given g ( x ) = 3 x 2 − 7 find
    1. g ( 3 t ) ,
    2. g ( t + 5 ) ,
    3. g ( 6 t − 4 ) ,
    4. g ( 4 x + 9 )
  5. Calculate f ( x + h ) when
    1. f ( x ) = x 2 ,
    2. f ( x ) = x 3 ,
    3. f ( x ) = 1 x . In each case write down the corresponding expression for f ( x + h ) − f ( x ) .
  6. If f ( x ) = 1 ( 1 − x ) 2 find f ( x ℓ ) .
  1. The argument is the input.
    1. 59,
    2. 19,
    3. − 1 ,
    4. 2.12.
    1. 2 x 2 + 4 ,
    2. 8 x 2 + 4 ,
    3. 2 x 2 + 4 ,
    4. 32 x 2 + 32 x + 12 ,
    5. 18 t 2 + 60 t + 54 ,
    6. 2 λ 2 + 4 ,
    7. 2 ( t − λ ) 2 + 4 ,
    8. 2 t 2 α 2 + 4 .
    1. 27 t 2 − 7 ,
    2. 3 t 2 + 30 t + 68 ,
    3. 108 t 2 − 144 t + 41 ,
    4. 48 x 2 + 216 x + 236 .
    1. x 2 + 2 x h + h 2 ,
    2. x 3 + 3 x 2 h + 3 x h 2 + h 3 ,
    3. 1 x + h .

      The corresponding expressions are
    1. 2 x h + h 2 ,
    2. 3 x 2 h + 3 x h 2 + h 3 ,
    3. 1 x + h − 1 x = − h x ( x + h ) .
  6. 1 ( 1 − x ℓ ) 2 .