Applications and Use of the Inverse Functions

Example 1

• Since logarithmic and exponential functions are inverses of each other, we can write the following.
  A = Log (B) if and only B = 10 ^A
  
• Use the above property of logarithmic and exponential functions to rewite the given equation as follows.
  x - 3 = 10 ²
  
• Solve for x to obtain.
  x = 103

Example 2

• We know that the range of a one to one function is the domain of its inverse. Let us first show that function f given above is a one to one function. Start with .
  f(a) = f(b)
  2 a / (a - 3) = 2 b / (b - 3)
  
• Multiply all terms of the above equation by (a - 3)(b - 3) and simplify to obtain.
  2a (b - 3) = 2 b(a - 3)
  
• Expand.
  2a b - 6 a = 2 ba - 6 b
  
• Add - 2 a b to both sides and simplify to obtain.
  a = b
  
• Hence the given function is a one to one. let us find its inverse.
  y = 2 x / (x - 3)
  
• Interchange x and y and solve for y.
  x = 2 y / (y - 3)
  y = 3x / (2 - x)
• The inverse of function f is given by.
  f^-1 (x) = 3x / (2 - x)
  
• The domain of f^-1 is the set of all real values except x = 2. Hence the range of f is the set of all real values except 2.

Example 3

• a) The opposite and the adjacent sides to angle t are x and 300 meters respectively, hence.
  tan(t) = x / 300
  
• We now use the property of the tangent function and its inverse.
  tan ^-1(tan(x)) = x
  
• To rewrite the equation tan(t) = x / 300 as follows.
  tan ^-1(tan(t)) = tan ^-1( x / 300 )
  
• Simplify the left side of the above equation to obtain.
  t = tan ^-1( x / 300 )
  
• b) The values of t at 150, 300 and 600 are found using a calculator..
  t(150) = 25.6 degrees (approximated to 1 decimal place)
  t(300) = 45.0 degrees
  t(600) = 63.4 degrees (approximated to 1 decimal place)
  
• c) We use the values of t in part b) and extra points and graph t as a function of x..
  

  x	t
  0	0
  150	25.6
  300	45.0
  600	63.4
  1200	76.0
  3000	84.3

  
  

Example 4

• The formula of the volume V of a right circular cone with height h and radius r is given by.
  V = (1/3) pi r² h
  
• Since we need to find the radius, we need to solve the above eqaution for r to obtain.
  r = SQRT(3 V / pi h)
  
• What we have done above is to find the inverse function of V. We do not need to interchange the variables V and r because they have different meaning in this problem..
  
• We now calculate the radius for each cone using the formula for the radius above..
  a) V = 200 , r = SQRT(3*200 / 50*pi) = 1.95 cm
  b) V = 400 , r = SQRT(3*400 / 50*pi) = 2.76 cm
  c) V = 800 , r = SQRT(3*800 / 50*pi) = 3.91 cm
  d) V = 1600 , r = SQRT(3*1600 / 50*pi) = 5.53 cm
  e) V = 3200 , r = SQRT(3*3200 / 50*pi) = 7.82 cm

Example 5

• We need to find t first by solving the given formula for t. Divide both sides of the given formula by 200,000..
  P / 200,000 = e ^{0.01 t}
  
• We now use the fact that the exponential and logarithmic functions are inverses of each other and rewrite the above exponential expression as follows.
  0.01 t = ln ( P / 200,000 )
  
• Solve for t..
  t = ln ( P / 200,000 ) / 0.01
• We now find t for the different values of P given above..
  a) P = 300,000 , t = ln ( 300,000 / 200,000 ) / 0.01 = 40.55 years , year 2041
  a) P = 400,000 , t = ln ( 400,000 / 200,000 ) / 0.01 = 69.31 years , year 2070
  a) P = 500,000 , t = ln ( 500,000 / 200,000 ) / 0.01 = 91.63 years , year 2092

More References and Links to the Inverse Functions