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1.1 Review of functions  (Page 3/28)

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( − ∞ , 0 ] = { x | x ≤ 0 } .

Here, the notation − ∞ refers to negative infinity, and it indicates that we are including all numbers less than or equal to zero, no matter how small. The set

( − ∞ , ∞ ) = { x | x is any real number }

refers to the set of all real numbers.

Some functions are defined using different equations for different parts of their domain. These types of functions are known as piecewise-defined functions . For example, suppose we want to define a function f with a domain that is the set of all real numbers such that f ( x ) = 3 x + 1 for x ≥ 2 and f ( x ) = x 2 for x < 2 . We denote this function by writing

f ( x ) = { 3 x + 1 x ≥ 2 x 2 x < 2 .

When evaluating this function for an input x , the equation to use depends on whether x ≥ 2 or x < 2 . For example, since 5 > 2 , we use the fact that f ( x ) = 3 x + 1 for x ≥ 2 and see that f ( 5 ) = 3 ( 5 ) + 1 = 16 . On the other hand, for x = −1 , we use the fact that f ( x ) = x 2 for x < 2 and see that f ( −1 ) = 1 .

Evaluating functions

For the function f ( x ) = 3 x 2 + 2 x − 1 , evaluate

  1. f ( −2 )
  2. f ( 2 )
  3. f ( a + h )

Substitute the given value for x in the formula for f ( x ) .

  1. f ( −2 ) = 3 ( −2 ) 2 + 2 ( −2 ) − 1 = 12 − 4 − 1 = 7
  2. f ( 2 ) = 3 ( 2 ) 2 + 2 2 − 1 = 6 + 2 2 − 1 = 5 + 2 2
  3. f ( a + h ) = 3 ( a + h ) 2 + 2 ( a + h ) − 1 = 3 ( a 2 + 2 a h + h 2 ) + 2 a + 2 h − 1 = 3 a 2 + 6 a h + 3 h 2 + 2 a + 2 h − 1
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For f ( x ) = x 2 − 3 x + 5 , evaluate f ( 1 ) and f ( a + h ) .

f ( 1 ) = 3 and f ( a + h ) = a 2 + 2 a h + h 2 − 3 a − 3 h + 5

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Finding domain and range

For each of the following functions, determine the i. domain and ii. range.

  1. f ( x ) = ( x − 4 ) 2 + 5
  2. f ( x ) = 3 x + 2 − 1
  3. f ( x ) = 3 x − 2
  1. Consider f ( x ) = ( x − 4 ) 2 + 5 .
    1. Since f ( x ) = ( x − 4 ) 2 + 5 is a real number for any real number x , the domain of f is the interval ( − ∞ , ∞ ) .
    2. Since ( x − 4 ) 2 ≥ 0 , we know f ( x ) = ( x − 4 ) 2 + 5 ≥ 5 . Therefore, the range must be a subset of { y | y ≥ 5 } . To show that every element in this set is in the range, we need to show that for a given y in that set, there is a real number x such that f ( x ) = ( x − 4 ) 2 + 5 = y . Solving this equation for x , we see that we need x such that
      ( x − 4 ) 2 = y − 5 .

      This equation is satisfied as long as there exists a real number x such that
      x − 4 = ± y − 5 .

      Since y ≥ 5 , the square root is well-defined. We conclude that for x = 4 ± y − 5 , f ( x ) = y , and therefore the range is { y | y ≥ 5 } .
  2. Consider f ( x ) = 3 x + 2 − 1 .
    1. To find the domain of f , we need the expression 3 x + 2 ≥ 0 . Solving this inequality, we conclude that the domain is { x | x ≥ −2 / 3 } .
    2. To find the range of f , we note that since 3 x + 2 ≥ 0 , f ( x ) = 3 x + 2 − 1 ≥ −1 . Therefore, the range of f must be a subset of the set { y | y ≥ −1 } . To show that every element in this set is in the range of f , we need to show that for all y in this set, there exists a real number x in the domain such that f ( x ) = y . Let y ≥ −1 . Then, f ( x ) = y if and only if
      3 x + 2 − 1 = y .

      Solving this equation for x , we see that x must solve the equation
      3 x + 2 = y + 1 .

      Since y ≥ −1 , such an x could exist. Squaring both sides of this equation, we have 3 x + 2 = ( y + 1 ) 2 .
      Therefore, we need
      3 x = ( y + 1 ) 2 − 2 ,

      which implies
      x = 1 3 ( y + 1 ) 2 − 2 3 .

      We just need to verify that x is in the domain of f . Since the domain of f consists of all real numbers greater than or equal to −2 / 3 , and
      1 3 ( y + 1 ) 2 − 2 3 ≥ − 2 3 ,

      there does exist an x in the domain of f . We conclude that the range of f is { y | y ≥ −1 } .
  3. Consider f ( x ) = 3 / ( x − 2 ) .
    1. Since 3 / ( x − 2 ) is defined when the denominator is nonzero, the domain is { x | x ≠ 2 } .
    2. To find the range of f , we need to find the values of y such that there exists a real number x in the domain with the property that
      3 x − 2 = y .

      Solving this equation for x , we find that
      x = 3 y + 2 .

      Therefore, as long as y ≠ 0 , there exists a real number x in the domain such that f ( x ) = y . Thus, the range is { y | y ≠ 0 } .
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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