Question 666709
<font face="Times New Roman" size="+2">


Since the inequality symbol is "strictly less than" rather than "less than or equal" your solution interval is going to be exclusive of the end points.  Therefore you need to use parentheses instead of brackets to surround your interval expression.  You have incorrectly identified the critical points.


Set the LHS equal to zero and solve the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 5x\ -\ 24\ =\ (x\ +\ 8)(x\ -\ 3)]


Hence *[tex \LARGE x\ =\ -8] or *[tex \LARGE x\ =\ 3]


so if you graph those two points on the *[tex \LARGE x] axis, you will have divided the *[tex \LARGE x] axis into three regions:  Negative infinity to -8, -8 to 3, and 3 to infinity.


The three intervals are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\infty,-8\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-8,3\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(3,\infty\right)]


All you have to do now is to select a value that is NOT an end point from each of the three intervals and find the value of your LHS function at each of those three selected values.  You will either get two positive and one negative value(s) or two negative and one positive value(s).  In the former case, your desired interval is just the center interval.  In the latter case, your desired interval is the union of the two outer intervals.  Just use a 'cup' symbol (a big 'U' or *[tex \Large \cup]) to indicate union


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>