Lesson Special advanced problems on finding the range of functions

Special advanced problems on finding the range of functions

Problem 1

(1)  The domain of this function is the set of all real numbers.

     In other words, the domain is the entire number line, from    to  .


     Question (1) is complete.



(2)  This function represents a parabola,

     Since the coefficient at x^1 is positive (it is 1 (one)), the parabola has the minimum.


     The range of this function is the set of its values, and it is the set of all real numbers 
     from the minimum of the parabola to infinity.

       
     So, to determine the range, we need to determine the minimum.


     The minimum of any quadratic function y = ax^2 + bx + c with positive coefficient "a" is at

          = .


     In our case it is   =  =  = -0.5.


     The minimum is   =  = 0.25 - 0.5 - 1 = -1.25.


     Therefore, the range of this quadratic function is from -1.25 to infinity.

     In the interval form, the range is  [-1.25,).


     Thus question (2) is complete.



(3)  To find x-intercepts, you need to solve this equation

         x^2 + x - 1 = 0.


     Use the quadratic formula

          =  = .


     The roots are   = =  0.618033989   (approximate),

               and   = = -1.618033989   (approximate).

Problem 2

Let real number "t" belongs to the range.  It means that

     = t    (1)

for some value of x.  Square bot sides of (1).  You will get

     = t^2

     = 0.    (2)


You can consider equation (2) as a quadratic equation relative .


The condition that it has a real solution for   is this inequality for the discriminant

     >= 0,  or   >= 0,  or  256 >= .


It implies   <= 4;  hence,  |t| <= 2,  or  -2 <= t <= 2.


Thus, equation (1) has a real solution if and only if  -2 <= t <= 2.


So, the range of the function    is this interval  -2 <= t <= 2,  or t belongs  [-2,2].    ANSWER

Problem 3

They ask about the range of this function  f(x) = 2x + .


The plot is shown under this link  

    https://www.desmos.com/calculator/lh7qjcdhic 

    https://www.desmos.com/calculator/lh7qjcdhic



The range is the set of real values t such that

    2x +  = t  for some x.


In other words, the range is the set of real numbers t such that the quadratic equation

    2x^2 + 3 = tx,  or  2x^2 - tx + 3 = 0  

has real solutions for x.


For it, the necessary and sufficient condition is that the discriminant is non-negative

    d = (-t)^2 - 4*2*3 >= 0,  or  t^2 >= 24,   or  |t| >= ,  or


          +------------------------------------+
          |    t <=    or  t >= .   |
          +------------------------------------+


Thus the range of the function  (fxg)  is the union of the two sets  


    (,] U [,).    ANSWER

Problem 4

Let real number  "t"  belongs to the range.  It means that

     = t    (1)

for some value of x.  Step by step, reduce equation (1) to standard form quadratic equation

     = ,

     = 0.    (2)


Equation (2) is a standard form quadratic equation   = 0  with coefficients

    a = 1,  b = -t,  c = 2-4t.


The condition that it has a real solution for   is this inequality for the discriminant


     >= 0,  or   >= 0,  or  t^2 + 16t -8 >= 0.


The roots of this quadratic equation for t are


     =  =  = -16.4853 (rounded)  and   =  =  = 0.4853 (rounded).


The discriminant is non-negative if and only if value of  "t"  is out of the roots' interval 


    t <=  = -16.4853  or  t >=  =0.48528.


Therefore, equation (1) has a real solution if and only if   < t <=    or   <= t < .


Thus the range of the function  f(x) = (x^2 + 2)/(x + 4) is the union of these intervals  (,]  and  [,).


ANSWER.  The range of the function    is the union of these intervals  (,]  and  [,).

Problem 5

Real number "t" belongs to the range of this function if and only if 

     = t for some real x.    (1)


Rewrite (1) in an equivalent form

     = t,

     = ,

     x^2 + 1 = tx^2 - 2tx + t,

     (1-t)x^2 + 2tx + (1-t) = 0.


This last quadratic equation has a real solution IF and ONLY IF the discriminant is non-negative

    d = b^2 - 4ac >= 0,

or

    (2t)^2 - 4*(1-t)*(1-t) >= 0,

    4t^2 - 4*(1-t)^2 >= 0,

    t^2 - 1 + 2t - t^2 >=0,

     2t >= 1,

      t >= 1/2.


At this point, the solution is complete.


ANSWER.  The domain is the set of all real numbers greater than or equal to 1/2.