Lesson Evaluating a function defined by functional equation

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Evaluating a function defined by functional equation


Problem 1

The function  f  satisfies   f%28sqrt%282x+-+1%29%29 = 1%2F%282x+-+1%29
for all  x  not equal to  1/2.   Find f(2).

Solution

        The idea is to find  x  such that   sqrt%282x-1%29 = 2
        and then calculate  f(2)  using the given functional equation.

        Below is an implementation of this idea in steps. 


Step 1.  We want to find x from equation

             sqrt%282x-1%29 = 2.


         Do all necessary transformations as follows

             2x-1 = 2^2 = 4,

             2x = 4 + 1 = 5,

              x = 5/2.



Step 2.  According to the functional equation,

                 f%28sqrt%282x+-+1%29%29 = 1%2F%282x+-+1%29.

         Substitute here x = 5/2.  Remember that x is determined in a way that sqrt%282x-1%29 = 2.

         Therefore, you will get    

                 f(2) = 1%2F%282%2A%285%2F2%29-1%29 = 1%2F%285-1%29 = 1%2F4.


At this point, the problem is solved completely.


ANSWER<.  f(2) = 1%2F4.

Problem 2

Suppose a function  f  is such that   f(1/x) - 3f(x) = x   for every non-zero  x.   Find  f(2).

Solution 

Write the given functional equation for x = 2 and for x = 1/2.


For x = 2, you will have

    f(1/2) - 3f(2) = 2              (1)


For x = 1/2, you will have

    f(1/(1/2)) - 3f(1/2) = 1/2,  

or

    f(2) - 3f(1/2) = 1/2.           (2)


Let  u = f(1/2),  v = f(2)  for breavity.  Then you can re-write (1) and (2) in the form

    u - 3v = 2                       (3)
    
    v - 3u = 1/2                     (4)


Now I want to solve the system of linear equations (3) and (4) and find v.


For it, I express  u = 2+ 3v  from (3) and substitute it into equation (4)

    v - 3*(2+3v) = 1/2.


In the last equation, multiply both sides by 2 to run from denominator

    2v - 6*(2+3v) = 1


Simplify and find v

    2v - 12 - 18v = 1

        -16v      = 1 + 12

        -16v      =   13

           v      =   -13%2F16.


Thus  v = f(2) = -13%2F16,  and the problem is just solved.


ANSWER.  f(2) = -13%2F16.

Problem 3

If   f%283x%2F%28x-4%29%29 = x%5E2+%2B+x+%2B+1,   what is the value of  f(5)?

Solution 

To solve the problem, find the value of x such that

    %283x%29%2F%28x-4%29 = 5.


It is easy: first, cross multiply to get

    3x = 5(x-4),


then simplify step by step

    3x = 5x - 20

    20 = 5x - 3x

    20 = 2x  --->  x = 20/2 = 10.


Now substitute x= 10 into the given formula. Since  %283x%29%2F%28x-4%29 = 5 at  x= 10,  you will get then

    f(5) = 10%5E2+%2B+10+%2B+1 = 100 + 10 + 1 = 111.


ANSWER.  Under imposed conditions,  f(5) = 111.

Problem 4

If   f(x) = 1/(1 - x) ,   find   (f(f(f(f...f)(sqrt2),  (45 times).

Solution 

If f(x) = 1%2F%281-x%29,  


then  f(f(x)) = 1%2F%281-1%2F%281-x%29%29 = %281-x%29%2F%28%281-x%29-1%29 = %28x-1%29%2Fx,


then  f(f(f(x))) = 1%2F%281-%28x-1%29%2Fx%29 = x%2F%28x-x%2B1%29 = x.


Thus, applying function f to any real number x =/= 1,  x =/= 0  three times, we get the value of  x  again.



In other words,  f(f(f(x))) == x identically, for all real x =/= 1,  x =/= 0.



So, for example,  f%28f%28f%28sqrt%282%29%29%29%29 = sqrt%282%29;  f%28f%28f%28sqrt%283%29%29%29%29 = sqrt%283%29;  f%28f%28f%28sqrt%285%29%29%29%29 = sqrt%285%29;  f%28f%28f%28sqrt%287%29%29%29%29 = sqrt%287%29;  

                  f%28f%28f%28root%283%2C2%29%29%29%29 = root%283%2C2%29;  f%28f%28f%28root%283%2C3%29%29%29%29 = root%283%2C3%29;  f%28f%28f%28root%283%2C5%29%29%29%29 = root%283%2C5%29;  f%28f%28f%28root%283%2C7%29%29%29%29 = root%283%2C7%29,  and so on.



Since 45 is a multiple of 3,  f applied to  sqrt%282%29  45 times is  sqrt%282%29;

                              f applied to  sqrt%283%29  45 times is  sqrt%283%29;

                              f applied to  sqrt%285%29  45 times is  sqrt%285%29;

                              f applied to  sqrt%287%29  45 times is  sqrt%287%29;


                              f applied to  root%283%2C2%29  45 times is  root%283%2C2%29%29;

                              f applied to  root%283%2C3%29  45 times is  root%283%2C3%29%29;

                              f applied to  root%283%2C5%29  45 times is  root%283%2C5%29%29;

                              f applied to  root%283%2C7%29  45 times is  root%283%2C7%29%29,

and so on.

Solved and significantly expanded.

For example,  f applied  2025  times to the number  root%282025%2C2025%29%29  is  root%282025%2C2025%29.

Similarly,  f  applied  2025  times to the number  2025%5E2025  is  2025%5E2025.

As well as  f  applied  2025  times to the number  2025!  is  2025! , again.

You can easily construct a million other examples.



My other lessons on Evaluating expressions in this site are
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