Examples with Solutions

Example 1

Solution to Example 1

• Let us first write the above function as an equation as follows
  
  
• solve the above function for x
  -x + 2 = ln (y)
  x = 2 - ln (y)
  
• x is a real number if y > 0 (argument of ln y must be positive). Hence the range of function f is given by
  y > 0 or the interval (0 , +∞)
  See graph of f below and examine the range graphically.
  

Matched Problem 1:

Example 2

Solution to Example 2

• Write the given function as an equation
  
  
• Solve the above equation for x
  x = (1 / 2)(ln (y - 3) -1)
  
• x is a real number for y - 3 > 0 (argument of ln (y - 3) must be positive). The range of the given function is then given by
  y > 3 or in interval form (3 , +∞)
  See graph of f below and examine the range graphically.
  

Matched Problem 2:

Example 3

Solution to Example 3

• Write the given function as an equation
  
  
  
• Solve the above for x to obtain
  x² = ln(y - 1)
  x = + or - √[ ln(y - 1) ]
  
• The above solutions are real if
  ln(y - 1) ≥ 0
  y - 1 ≥ 1
  y ≥ 2
  
• Hence the range of the given function is given by
  y ≥ 2 or in interval form [ 2 , + ∞ )

Matched Problem 3:

Example 4

Solution to Example 4

• We first write the given function as an equation as follows
• Solve the above for x
  y - 3 = -2 e^-x2
  e^-x2 = (y - 3) / (-2)
  -x² = ln [ (y - 3) / (-2) ]
  x = + or - √( - ln [ (y - 3) / (-2) ])
  
• x is real if the argument of ln is positive and the radicand is positive or zero. Hence the following inequalities
  (y - 3) / (-2) > 0 and - ln [ (y - 3) / (-2) ] ≥ 0
  the solution set of (y - 3) / (-2) > 0 is y < 3
  the solution set of - ln [ (y - 3) / (-2) ] ≥ 0 is given by (y - 3) / (-2) ≤ 1 which gives y ≥ 1
  
• the range of f is given by
  1 ≤ y < 3 or in interval form [ 1 , 3 )

Matched Problem 4:

Answers to Matched Problems

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