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

I corrected my former answer.

 




There are vertical asymptotes at ,
and there are no points between those two vertical asymptotes.

We know that there is a vertical asymptote at  so y goes to infinity on the right.  

Also since the limit of y as x approaches +infinity is 1, there is a horizontal
asymptote y = 1 and y is always above that horizontal asymptote when


Also since the limit of y as x approaches -infinity is -1, there is a horizontal
asymptote y = -1 and y is always above that horizontal asymptote when











Factor the negative out of the last factor:



The discriminant is not negative:



Since  is always true, we must have




Since y is always negative left of its left asymptote



However y is always above its horizontal asymptote y = 1 to the right of its
vertical asymptote 

Since the denominator is always positive, y is always negative when x<-1 

So on the left of the vertical asympote.  

However on the right of its horizontal asymptote  y is always
above its horizontal asymptote y = 1, so y > 1 when 

So the range of the function is



or you can rationalize the denominator:



Edwin

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

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


In the area x > ,  when x approaches to  from the right side, the function 
asymptotically goes to infinity.


In the same area, x > , when x goes to positive infinity, the function is asymptotically 
close to  ,  so, it goes to 1 unit monotonically decreasing.


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



In the area x < ,  when x approaches to  from the left side, the function 
asymptotically goes to negative infinity.


In the same area, x < , when x goes to negative infinity, the function is asymptotically 
close to  ,  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
 < 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 .


    +------------------------------------------------------------------------+
    |   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   .  Let's write  

     = t,    (1)

by introducing new variable 't'.

We want to find maximum value of 't' in the area   < x < .


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 =  = 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 = +/-  = +/-  = +/-  = +/- 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 = 
for its left branch over the domain   < x < -3.


Thus the range of the given function is  ( ,  ] U ( , ),
the union of two intervals.


It is consistent with my plot, referred above.