Question 1208034
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Find the range of f(x) = (x^2 + 2)/(x + 4) algebraically.
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<pre>
Let real number  "t"  belongs to the range.  It means that

    {{{(x^2+2)/(x+4)}}} = t    (1)

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

    {{{x^2 + 2}}} = {{{t*(x+4)}}},

    {{{x^2 - tx + (2-4t)}}} = 0.    (2)


Equation (2) is a standard form quadratic equation  {{{ax^2 + bx + c}}} = 0  with coefficients

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


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


    {{{b^2 - 4ac}}} >= 0,  or  {{{(-t)^2 - 4*1*(2-4t)}}} >= 0,  or  t^2 + 16t -8 >= 0.


The roots of this quadratic equation for t are


    {{{t[1]}}} = {{{(-16 - sqrt(288))/2}}} = {{{-8 - 6*sqrt(2)}}} = -16.4853 (rounded)  and  {{{t[2]}}} = {{{(-16 + sqrt(288))/2}}} = {{{-8 + 6*sqrt(2)}}} = 0.4853 (rounded).


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


    t <= {{{-8 - 6*sqrt(2)}}} = -16.4853  or  t >= {{{-8 + 6*sqrt(2)}}} =0.48528.


Therefore, equation (1) has a real solution if and only if  {{{-infinity}}} < t <=  {{{-8 - 6*sqrt(2)}}}  or  {{{-8 + 6*sqrt(2)}}} <= t < {{{infinity}}}.


Thus the range of the function  f(x) = (x^2 + 2)/(x + 4) is the union of these intervals  ({{{-infinity}}},{{{-8 - 6*sqrt(2)}}}]  and  [{{{-8 + 6*sqrt(2)}}},{{{infinity}}}).


<U>ANSWER</U>.  The range of the function  {{{(x^2 + 2)/(x + 4)}}}  is the union of these intervals  ({{{-infinity}}},{{{-8 - 6*sqrt(2)}}}]  and  [{{{-8 + 6*sqrt(2)}}},{{{infinity}}}).
</pre>

Solved.


It is how to solve this problem using only elementary Algebra and without using Calculus.



This problem is a typical Math Olympiad level problem or Math circle level problem 
for 9-th grades high school students, who just know Algebra, but still don't know Calculus.