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