Question 1203358
.
Find the range of f(x)= sqrt(x^2-2) + sqrt((1+x^2)/(2+x^2)) - (8-x^2)/(x^2-4x-1)
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<pre>
Our function f(x) is the sum of three addends p(x) = {{{sqrt(x^2-2)}}},  q(x) = {{{sqrt((1+x^2)/(2+x^2))}}}  and  r(x) = {{{-(8-x^2)/(x^2-4x-1)}}}.  

Their plots are shown in figures 1, 2 and 3.


    {{{graph( 400, 400, -10, 10, -10, 10,
          sqrt(x^2-2)
)}}}

       Figure 1. Plot p(x) = {{{sqrt(x^2-2)}}}



    {{{graph( 400, 400, -10, 10, -5, 5,
          sqrt((1+x^2)/(2+x^2))
)}}}

       Figure 2. Plot q(x) = {{{sqrt((1+x^2)/(2+x^2))}}}



    {{{graph( 400, 400, -10, 10, -10, 10,
          -(8-x^2)/(x^2-4x-1)
)}}}

       Figure 3. Plot r(x) = {{{-(8-x^2)/(x^2-4x-1)}}}



    {{{graph( 400, 400, -10, 10, -10, 10,
          sqrt(x^2-2) sqrt((1+x^2)/(2+x^2)) -(8-x^2)/(x^2-4x-1)
)}}}


   Figure 4. Plot f(x) = {{{sqrt(x^2-2)}}} + {{{sqrt((1+x^2)/(2+x^2))}}} - {{{(8-x^2)/(x^2-4x-1)}}}


From Figure 1 and from the formula for p(x), is is easily seen that p(x) is defined outside of  { {{{-sqrt(2)}}} < x < {{{sqrt(2)}}} },
and so (hence) f(x) is also defined outside of this interval.


On the left of this interval, the range of p(x) is unbounded from 0 to infinity in positive direction.


From Figure 2 and from the formula for q(x), it is clear that function q(x) is defined on the whole number line 
and is continuous function bounded in infinity (both in positive and negative infinity), so this function 
is bounded continuous function.


From Figure 3 and from the formula for r(x), it is clear that the range of function r(x) is unbounded in negative 
direction in vicinity of its pole, which is located at the zero of its denominator x = {{{(4+sqrt(4^2+4))/2}}} = {{{2 + sqrt(5)}}} = 4.24.... *)
At the same time, as x goes to positive or negative infinity, function r(x) remains bounded.


Collecting all these properties and making a plot of the function f(x) in Figure 4, we see that
all these listed properties make the range of function f(x) as whole number line without any holes.
</pre>

Solved.


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*) &nbsp;&nbsp;This pole &nbsp;&nbsp;x = {{{2 + sqrt(5)}}} = 4.24.... &nbsp;does make a hole in the domain 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the function &nbsp;f(x), &nbsp;but does not make a hole in its range.