Question 1210370
.
Using the discriminant method, find the range of the function:
y = (x+1)/(sqrt(x² - 5))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



        I will solve this problem and will find the range of the given function algebraically, 

        using the discriminant method.


        But for better understanding, I will start from the plot, which I prepared using

        the plotting tool DESMOS. This plotting tool is free of charge at this website

        http://desmos.com/calculator/


        The link to my plot is this https://www.desmos.com/calculator/lhfouvblry



<pre>
Looking at the function and at the plot, we see that the domain of the function is  

    x^2-5 > 0,  or  x < {{{-sqrt(5)}}} U x > {{{sqrt(5)}}}   (the union of two sets).


In the area x > {{{sqrt(5)}}},  when x approaches to {{{sqrt(5)}}} from the right side, the function 
asymptotically goes to infinity.


In the same area, x > {{{sqrt(5)}}}, when x goes to positive infinity, the function is asymptotically 
close to  {{{(x+1)/x}}},  so, it goes to 1 unit monotonically decreasing.


Hence, the range of the right branch of the function at positive x > {{{sqrt(5)}}}  is  from 1 to infinity, 
or ( {{{1}}},{{{infinity}}} ).



In the area x < {{{-sqrt(5)}}},  when x approaches to {{{-sqrt(5)}}} from the left side, the function 
asymptotically goes to negative infinity.


In the same area, x < {{{-sqrt(5)}}}, when x goes to negative infinity, the function is asymptotically 
close to  {{{(x+1)/abs(x)}}},  so, it goes to negative 1 unit, or (-1), monotonically decreasing.


But at x = -3, the value of the function is  (-1), as it is easy to check, and then, in the area
{{{-infinity}}} < x  < 3, the function is greater than -1 and tends to -1 from greater values than -1.


Hence, the function has a local maximum somewhere between -3 and {{{-infinity}}}.


    +------------------------------------------------------------------------+
    |   The major goal of this problem is to determine this local maximum.   |
    |   As we get this local maximum, we will solve the problem completely.  |
    +----------------------------------------------------------------------=-+


OK.  The next part of the solution is the   {{{highlight(highlight(CULMINATION))}}}.  Let's write  

    {{{(x+1)/(sqrt(x^2 - 5))}}} = t,    (1)

by introducing new variable 't'.

We want to find maximum value of 't' in the area  {{{-infinity}}} < x < {{{-sqrt(5)}}}.


From equation (1), we have

    (x+1)^2 = t^2*(x^2-5),

    x^2 + 2x + 1 = t^2*x^2 - 5t^2,

    (t^2-1)x^2 - 2x - (5t^2+1) = 0.    (2)


    +---------------------------------------------------------------------+
    |    The discriminant principle says that the extremal value of 't'   |
    |    in this equation is achieved  when the discriminant              |
    |                  of this equation is zero.                          |
    |    It is the condition that equation (2) has emerging roots in x.   |
    +---------------------------------------------------------------------+


So, the discriminant of equation (2) is

    d = b^2 - 4ac = (-2)^2 + 4*(t^2-1)*((5t^2+1) = 4 + 4(t^2-1)*(5t^2+1).


The equation d = 0 is then

    4 + 4(t^2-1)*(5t^2+1) = 0,

or

    (t^2-1)*(5t^2+1) = -1.    (3)


For simplicity, let's introduce new variable  u = t^2.   

Then equation  (3)  takes the form

    (u-1)*(5u+1) = -1,

    5u^2 - 4u - 1 = -1,

    5u^2 - 4u = 0,

     u*(5u - 4) = 0.


It has the roots  u = 0  and  u = {{{4/5}}} = 0.8.


The root u = 0  is irrelevant and we reject it.


The root u = 4/5 = 0.8  gives us

    t^2 = = 0.8,  t = +/- {{{sqrt(0.8)}}} = +/- {{{sqrt(4/5)}}} = +/- {{{2/sqrt(5)}}} = +/- 0.894427.


The positive value is irrelevant (extraneous), and we reject it.

But the negative value, t = -0.894427  gives us the maximum value of the function  y = {{{(x+1)/(sqrt(x^2 - 5))}}}
for its left branch over the domain  {{{-infinity}}} < x < -3.


Thus the range of the given function is  ( {{{-infinity}}}, {{{-2/sqrt(5)}}} ] U ( {{{1,}}},{{{infinity}}} ),
the union of two intervals.


It is consistent with my plot, referred above.
</pre>

At this point, the problem is solved completely.