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Algebraic functions

By allowing for quotients and fractional powers in polynomial functions, we create a larger class of functions. An algebraic function    is one that involves addition, subtraction, multiplication, division, rational powers, and roots. Two types of algebraic functions are rational functions and root functions.

Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. In particular, a rational function    is any function of the form f ( x ) = p ( x ) / q ( x ) , where p ( x ) and q ( x ) are polynomials. For example,

f ( x ) = 3 x 1 5 x + 2 and g ( x ) = 4 x 2 + 1

are rational functions. A root function    is a power function of the form f ( x ) = x 1 / n , where n is a positive integer greater than one. For example, f ( x ) = x 1 / 2 = x is the square-root function and g ( x ) = x 1 / 3 = x 3 is the cube-root function. By allowing for compositions of root functions and rational functions, we can create other algebraic functions. For example, f ( x ) = 4 x 2 is an algebraic function.

Finding domain and range for algebraic functions

For each of the following functions, find the domain and range.

  1. f ( x ) = 3 x 1 5 x + 2
  2. f ( x ) = 4 x 2
  1. It is not possible to divide by zero, so the domain is the set of real numbers x such that x 2 / 5 . To find the range, we need to find the values y for which there exists a real number x such that
    y = 3 x 1 5 x + 2 .

    When we multiply both sides of this equation by 5 x + 2 , we see that x must satisfy the equation
    5 x y + 2 y = 3 x 1 .

    From this equation, we can see that x must satisfy

    2 y + 1 = x ( 3 5 y ) .

    If y = 3 / 5 , this equation has no solution. On the other hand, as long as y 3 / 5 ,
    x = 2 y + 1 3 5 y

    satisfies this equation. We can conclude that the range of f is { y | y 3 / 5 } .
  2. To find the domain of f , we need 4 x 2 0 . When we factor, we write 4 x 2 = ( 2 x ) ( 2 + x ) 0 . This inequality holds if and only if both terms are positive or both terms are negative. For both terms to be positive, we need to find x such that
    2 x 0 and 2 + x 0 .

    These two inequalities reduce to 2 x and x −2 . Therefore, the set { x | 2 x 2 } must be part of the domain. For both terms to be negative, we need
    2 x 0 and 2 + x 0 .

    These two inequalities also reduce to 2 x and x −2 . There are no values of x that satisfy both of these inequalities. Thus, we can conclude the domain of this function is { x | 2 x 2 } .
    If −2 x 2 , then 0 4 x 2 4 . Therefore, 0 4 x 2 2 , and the range of f is { y | 0 y 2 } .
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Find the domain and range for the function f ( x ) = ( 5 x + 2 ) / ( 2 x 1 ) .

The domain is the set of real numbers x such that x 1 / 2 . The range is the set { y | y 5 / 2 } .

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The root functions f ( x ) = x 1 / n have defining characteristics depending on whether n is odd or even. For all even integers n 2 , the domain of f ( x ) = x 1 / n is the interval [ 0 , ) . For all odd integers n 1 , the domain of f ( x ) = x 1 / n is the set of all real numbers. Since x 1 / n = ( x ) 1 / n for odd integers n , f ( x ) = x 1 / n is an odd function if n is odd. See the graphs of root functions for different values of n in [link] .

An image of two graphs. The first graph is labeled “a” and has an x axis that runs from -2 to 9 and a y axis that runs from -4 to 4. The first graph is of two functions. The first function is “f(x) = square root of x”, which is a curved function that begins at the origin and increases. The second function is “f(x) = x to the 4th root”, which is a curved function that begins at the origin and increases, but increases at a slower rate than the first function. The second graph is labeled “b” and has an x axis that runs from -8 to 8 and a y axis that runs from -4 to 4. The second graph is of two functions. The first function is “f(x) = cube root of x”, which is a curved function that increases until the origin, becomes vertical at the origin, and then increases again after the origin. The second function is “f(x) = x to the 5th root”, which is a curved function that increases until the origin, becomes vertical at the origin, and then increases again after the origin, but increases at a slower rate than the first function.
(a) If n is even, the domain of f ( x ) = x n is [ 0 , ) . (b) If n is odd, the domain of f ( x ) = x n is ( −∞ , ) and the function f ( x ) = x n is an odd function.

Finding domains for algebraic functions

For each of the following functions, determine the domain of the function.

  1. f ( x ) = 3 x 2 1
  2. f ( x ) = 2 x + 5 3 x 2 + 4
  3. f ( x ) = 4 3 x
  4. f ( x ) = 2 x 1 3
  1. You cannot divide by zero, so the domain is the set of values x such that x 2 1 0 . Therefore, the domain is { x | x ± 1 } .
  2. You need to determine the values of x for which the denominator is zero. Since 3 x 2 + 4 4 for all real numbers x , the denominator is never zero. Therefore, the domain is ( −∞ , ) .
  3. Since the square root of a negative number is not a real number, the domain is the set of values x for which 4 3 x 0 . Therefore, the domain is { x | x 4 / 3 } .
  4. The cube root is defined for all real numbers, so the domain is the interval (−∞, ∞).
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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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