Re-write so that the inequality can be decomposed into a piece-wise function:
This inequality has two critical points;
Hence there are three intervals to consider when creating a piece-wise function definition:
Since
So we need to find the interval that satisfies both
But the intervals
Next, for values in
So, set the function less than zero and solve:
Hence, the function is true on the interval
Finally, for values in
So the inequality holds for values in the interval
That leaves the value
So now we can specify the union of the two valid intervals which include the
Is the complete solution set interval.
Compare this result to the portion of the graph of
John
My calculator said it, I believe it, that settles it