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\(For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ \)

Guest Apr 6, 2017
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2+0 Answers

 #1
avatar+75302 
+1

 

This one is a little difficult.....however.......note that x^2 + x + 3  is positive for all x

 

We can see this because this is a parabola that turns upward and the vertex is  (-1/2, 2.75)  .....which is above the x  axis......so this part of the function is never negative

 

So.....only the denominator can  make this function negative....and we have

 

 [ 2x^2 + x - 6 ] ≥ 0

 

Factor the denominator

 

(2x -3) ( x + 2)  ≥ 0

 

Setting each factor to  0 and solving for x, we have that   x = 3/2   and x =  -2

 

And we can see that when    -2 < x < 3/2   the denominator will be  negative

 

Then...the function will be   ≥ 0  on these intervals :

 

 (-infinity , -2) U (3/2, +infinity)

 

Notice that we have to exclude  -2 and 3/2  from the soution because these would make the denominator 0

 

See the graph here :   https://www.desmos.com/calculator/dztw4t5o0m

 

 

cool cool cool

CPhill  Apr 6, 2017
 #2
avatar+1148 
+1

Very nice Cphill!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

tertre  Apr 7, 2017

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