Find Range of Functions

Find the range of real valued functions using different techniques.

Example 1: Find the Range of function f defined by

f (x) = -3

Solution to Example 1

  • The given function has a constant value 3 and therefore the range is the set

    {3}

Example 2: Find the Range of function f defined by

f (x) = 4 x + 5

Solution to Example 2

  • Assuming that the domain of the given function is the set of all real numbers R, so that the variable x takes all values in the interval

    (-∞ , +∞)

    If x takes all values in the interval (-∞ , +∞) then 4 x + 5 takes all values in the interval (-∞ , +∞) and the range of the given function is given by the interval (-∞ , +∞)


Example 3: Find the Range of function f defined by

f (x) = x 2 + 5

Solution to Example 3

  • Assuming that the domain of the given function is R meaning that x takes all values in the interval (-∞ , +∞) which means that x 2 is either zero or positive. Hence we can write the following inequality

    x 2 ≥ 0

    Add 5 to both sides of the inequality to obtain the inequality

    x 2 + 5 ≥ = 0 + 5 or f(x) ≥ 5

    The range of f (x) = x 2 + 5 is given by the interval: [5 , +∞)


Example 4: Find the Range of function f defined by

f (x) = -2 x 2 + 4x - 7

Solution to Example 4

  • We first write the given quadratic function in vertex form by completing the square
    f (x) = -2 x 2 + 4x - 7 = -2(x 2 - 2 x) - 7 = -2( (x - 1) 2 - 1) - 7 = -2(x - 1) 2 - 5
  • Assuming the the domain of the given function is R with x taking any value in the interval, (x - 1) 2 is either zero or positive. Hence we start by writing the inequality
    (x - 1) 2 ≥ 0

  • Multiply both sides of the inequality by - 2 and change the symbol of inequality to obtain
    - 2 (x - 1) 2 ≤ 0

  • Add - 5 to both sides of the inequality to obtain

    - 2 (x - 1) 2 - 5≤ - 5 or f(x) ≤ - 5 and hence the range of function f is given by the interval (-∞ - 5]


Example 5: Find the Range of function f defined by

f (x) = -3 e2x + 5 + 2

Solution to Example 5

  • Assuming that the domain of the given function is R and therefore x takes all values in the interval (-∞ , +∞) and therefore the exponent 2x + 5 takes all values in the interval (-∞ , +∞), we can therefore write the following inequality (basic exponential function always positive)
    e2x + 5 > 0

  • Multiply both sides of the inequality by -3 and change symbol of inequality to obtain

    - 3 e2x + 5 < 0

  • Add 2 to both sides of the inequality to obtain

    - 3 e2x + 5 + 2 < 2 or f(x) < 2

    The range of the given function is given by the interval (-∞ , 2).


Example 6: Find the Range of function f defined by

f (x) = -3 ex 2 + 5 + 2

Solution to Example 6

  • Assuming the domain of the given function is R, we can start by writing the inequality
    x 2 ≥ 0

  • Add 5 to both sides of the inequality
    x 2 + 5≥ 5

  • The basic exponential function being an increasing function, we can use the above to write the inequality

    e x 2 + 5 ≥ e 5

  • Multiply both sides of the inequality by -3 and add 2 to both sides to obtain

    - 3 e x 2 + 5 + 2 ≤ -3 e 5 + 2

  • Note the the left hand side of the inequality is equal to f(x). Hence
    f(x)< -3 e 5 + 2 which means that the range of function f is given by the interval: (-∞ , -3 e 5 + 2]


Example 7: Find the Range of function f defined by

f (x) = (x - 1) / (x + 2)

Solution to Example 7

  • For this rational function, a direct algebraic method similar to those above is not obvious. Let us first find its inverse, the domain of its inverse which give the range of f.
  • We first prove that f is a one to one function and then find its inverse. For a function to be a one to one, we need to show that
    If f(a) = f(b) then a = b.

    (a - 1) / (a + 2) = (b - 1) / (b + 2)

  • Cross multiply, expand and simplify

    (a - 1)*(b + 2) = (b - 1)*(a + 2)

    a b + 2 a - b - 2 = a b + 2b - a - 2

    3a = 3b , which finally gives a = b and proves that f is a one to one function.

  • Let us find the inverse of f
    y = (x - 1) / (x + 2)

  • Solve for x

    x = (2y + 1) / (1 - y)

  • change y into x and x into y and write the inverse function

    f -1 (x) = y = (2x + 1) / (1 - x)

  • The range of f is given by the domain of f -1 and is therefore given by the interval

    (-∞ , 1) U (1 , + ∞)


Example 8: Find the Range of function f defined by

f (x) = -3 ln (x + 3) - 2

Solution to Example 8

  • The range of ln (x) is given by the interval (-∞ , + ∞). Since the graph of ln(x + 3) is the graph of ln (x) shifted 3 units to the left, the range of ln(x + 3) is also given by the interval (-∞ , + ∞). The graph of -3 ln(x + 3) is that of ln (x + 3) reflected on the x-axis, because of the minus sign and expanded vertically by 3 and therefore the range is still given by the interval (-∞ , + ∞). The graph of f is that of -3 ln(x +3) shifted 2 units up and therefore the range is also given by the interval (-∞ , + ∞).


Example 9: Find the Range of function f defined by

f (x) = 1 / (x 2 + 4)

Solution to Example 9

  • x 2 is a quantity which may be zero or positive and therefore we can write

    x 2 ≥ 0 which also gives x 2 + 4≥ 4

  • Divide both side of the inequality x 2 + 4 ≥ 4 by the positive quantity 4(x 2 + 4) to obtain

    1/4 ≥ 1 / (x 2 + 4)

  • The right side of the inequality is equal to f(x). Hence the above inequality may be written as

    f(x) ≤ 1/4

  • Because 1 / (x 2 + 4) is always positive and never zero but may be very close to zero as x increases or decreases indefinitely , the range of f is given by the interval (0; 1/4]


Example 10: Find the Range of function f defined by

f (x) = sqrt(x 2 + 4x + 4) - 6

Solution to Example 10

  • Note that x 2 + 4x + 4 = (x + 2) 2. Therefore

    f (x) = sqrt(x 2 + 4x + 4) - 6 = sqrt( (x + 2) 2 ) - 6 = |x + 2| - 6

  • The range of |x + 2| is given by the interval [0 , +∞). The graph of f is that of |x + 2| shifted down by 6 units and therefore the range of is given by the interval [-6 , +∞)


Example 10 Find the Range of function f defined by

f (x) = -2 sin(3x - Pi) + 1.5

Solution to Example 11

  • From basic trigonometry we know that the range of values of sine function is [-1 , 1]. Hence

    -1 ≤ sin(3x - Pi) ≤ 1

  • Multiply all terms by -2; change symbols of inequalities and add 1.5 to all terms

    2 + 1.5 ≥ -2 sin(3x - Pi) + 1.5 ≥ -2 + 1.5

  • Simplify and rewrite as

    -0.5 ≤ f(x) ≤3.5 which gives the range of f as the interval [-0.5 , 3.5].


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