Module 4

Module 4.1- Exponential Functions Examples Example 1- The court-imposed fine for the city of Yonkers You may recall that the city council defied a district court-order to desegregate housing in Yonkers. So, Judge Sand imposed a fine against the city starting at $100 on August 2nd and doubling daily until officials obeyed the court order. Previously we modeled the relationship between the number of days since August 2 nd  (when the court-order was issued) and the the fine for the city of Yonkers. Here again is our model. y = 100 ⋅ 2 t Let's identify how our model fits the definition of an exponential function given above.  t = ¿  number of days since August 2 nd  and this is  x  in  f ( x )= a ⋅ b x .  y = ¿  the fine (in dollars) imposed on the city of Yonkers and this is  f ( x )  in  f ( x )= a ⋅ b x .  The initial value is  a = $ 100 . And, the conditions that  a  is a real number and  a≠ 0   are met.  The growth factor is b = 2 . And the conditions that  b  is a positive real number and  b≠ 1  are met. Example 2- Repopulating a wildlife refuge Prior to 2012, the deer population in a wildlife refuge area had been hunted to elimination. In 2014, 80 deer were reintroduced into the wildlife refuge area.  By 2020, the population had grown to 180 deer.  If this population grows exponentially, find an exponential model for the deer population in this wildlife refuge area and use it to predict the deer population in 2025. Solution: We define our input variable to be  t , the number of years after 2014, and the output variable to be  N , the number of deer in the wildlife refuge area  t years after 2014. Then our input-output pairs must have the form  ( t ,N ) .   In 2014, t = 0  and  N = 80  . So, the first input-output pair is  ( 0,80 ) . In 2020,  t = 6  and  N = 180 . So, the second input-output pair is  ( 6,180 ) . Since we are given that the deer population is growing exponentially, we will use the model  N ( t )= a ⋅ b t . To find the model for the deer population, we need to find values for parameter  a  (the initial value) and the parameter  b  (the growth factor). 

Notice that by choosing our input variable  t to be the number of years after 2014, we "gave" ourselves the initial value for the function. So, we know that the parameter  a = 80 .  So, we can substitute  a = 80  into  N ( t )= a ⋅ b t . N ( t )= 80 ⋅ b t We have used the first input-output pair to obtain the parameter  a . Now, to find the parameter  b  we'll substitute the second input-output pair into  N ( t )= 80 ⋅ b t . The second input-output pair is  ( 6,180 ) . This tells us that  t = 6 , and  N ( 6 )= 180 . Making the substitutions gives us ... 180 = 80 ⋅ b 6 Dividing both sides by 80, gives us ... b 6 = 180 80 Taking the 6th root of both sides, gives us ... b = 6 √ 180 80 = ( 180 80 ) ( 1 6 ) Entering the expression  ( 180 80 ) ( 1 6 ) into our graphing calculators gives us ... b≈ 1.1447 So the deer population is modeled by the exponential function  N ( t )= 80 ( 1.1447 ) t . In 2025,  t = 2025 − 2014 = 11 . N ( 11 )= 80 ( 1.1447 ) 11 ≈ 354 . In 2025 we predict that population of deer in the wildlife refuge area will be 354. Example 3- The growth rate for the population of deer Find the growth rate  for the population of deer from the previous example. Then interpret the growth rate in context. Solution: Recall that the growth factor is  b = 1 + r  where  r  is the growth rate in decimal form.  Substituting  b≈ 1.1447  into  b = 1 + r  and solving for  r , gives us  r = 0.1447 . This is the growth rate in decimal form.

So the grow rate is 14.47%. This means that the deer population in the wildlife refuge area is increasing 14.47% each year.   Example 4- Find the exponential function passing through two points Find a formula for an exponential function passing through the points  (− 2,6 )  and  ( 2,1 ) . Solution: Since we don’t have the initial value, we will take a general approach that will work for any function form with unknown parameters. We will substitute both of the given input- output pairs into the function form  y = a ⋅ b x   solve for the unknown parameters,  a  and  b . All input-output pairs have the form  ( x , y )  . Equation 1: Substituting  (− 2,6 )  into  y = a ⋅ b x  gives us  6 = a ⋅ b − 2 . Equation 2: Substituting  ( 2,1 )  into  y = a ⋅ b x  gives us  1 = a ⋅ b 2 . Equation 1:    6 = a ⋅ b − 2 Equation 2:   1 = a ⋅ b 2 First we solve  Equation 1   for  a . Multiplying both sides by  b 2  yields  a = 6 b 2 . Substituting  a = 6 b 2  into  Equation 2  yields  1 =( 6 b 2 ) ⋅ b 2 . Now, we just solve for  b . 1 =( 6 b 2 ) ⋅ b 2 1 = 6 b 4 b 4 = 1 6 b = 4 √ 1 6 (Note: since we will use  b  to find  a , we avoid round-off error in our calculations by not rounding  b  yet.) To find  a , we substitute  b = 4 √ 1 6  into  a = 6 b 2 .   a = 6 ( 4 √ 1 6 ) 2 = 6 ⋅ √ 1 6 Now, we are ready to round.