Examples with Solutions

Example 1

Solution to Example 1

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

Example 2

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

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

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
  
• The domain of the given function is R with x taking any value in the interval (-∞ , +∞) hence (x - 1) ² is either zero or positive. 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

Solution to Example 5

• The domain of the given function is R and therefore x takes all values in the interval (-∞ , +∞). The exponent 2x + 5 takes all values in the interval (-∞ , +∞), we can therefore write the following inequality (basic exponential function always positive)
  e^{2x + 5} > 0
  
• Multiply both sides of the inequality by -3 and change symbol of inequality to obtain
  
  - 3 e^{2x + 5} < 0
  
  
• Add 2 to both sides of the inequality to obtain
  
  - 3 e^{2x + 5} + 2 < 2 or f(x) < 2
  
  The range of the given function is given by the interval (-∞ , 2).

Example 6

Solution to Example 6

• 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

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

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

Solution to Example 9

• x ² is a quantity which may be zero or positive and therefore we can write
  x ² ≥ 0
• add 4 to both sides of the inequality to obtain
  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

Solution to Example 10

• Note that x ² + 4x + 4 = (x + 2) ². Therefore
  f (x) = √(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 11

Solution to Example 11

• From basic trigonometry we know that the range of values of sine function is [-1 , 1]. Hence
  -1 ≤ sin(3x - π) ≤ 1
  
• Multiply all terms by -2; change symbols of inequalities and add 1.5 to all terms
  2 + 1.5 ≥ -2 sin(3x - π) + 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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