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April 3rd, 2014, 07:58 AM  #1 
Newbie Joined: Apr 2014 From: The Netherlands Posts: 7 Thanks: 2  Easy inequality question
Hi, I think the answer to this is pretty simple. Assume we have the function f(x) = 3x² + px + 3 Now the question is: for which p does f(x) have a negative minimum? So, we know that the discriminant must be greater than zero. D (b²  4ac) > 0 p²  36 > 0 p² > 36 Now, this is where I get confused, we derive 6 and 6, so I would write: p > 6 and p > 6 I know that this is wrong, but I still don't get why the sign changes, can someone help me out? Here is another, similar one. f(x) = px² + 2px + 3 for which p does f(x) have a negative minimum? Again, D > 0 4p²  12p > 0 4p (p  3) > 0 So, we're at that point again. I'd write p > 0 and p > 3, but that's not correct. (I know p < 0 will ultimately cancel out aswell because in that case the first term would be negative but we are looking for a minimum). Thanks in advance 
April 3rd, 2014, 08:49 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  This guarantees that the function has real roots. To have a negative minimum also requires that a>0. You need to be more careful about what you get from this. Not $p \gt 6, \, p \gt 6$ since $p = 0 \gt 6$ is clearly not a solution. But if $p = 7$ we find that $p^2 = 49 \gt 36$ is a solution. What we really need, therefore, is $p \gt 6$ Again, we do not get $p \gt 0, \, p \gt 3$ from this. If $p = 1 \gt 0$ we get $4p > 0, \, p  3 < 0$ so $4p (p  3) \lt 0$ and so $p = 1$ is not a solution. But $p = 1$ gives $4p \lt 0, p3 \lt 0$ so $4p (p  3) \gt 0$ which is a solution. Our answer is therefore, that all $p$ except for $0 \le p \le 3$ give real roots, but when $p \lt 0$ the curve has no negative minimum since it goes to $\infty$ as $x \rightarrow \pm \infty$. Thus $p > 3$ gives us negative minima. 
April 3rd, 2014, 08:56 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
For the first one, you have gotten: $\displaystyle p^2>36$ At this point, you would write: $\displaystyle p>6$ which means $p$ is in: $\displaystyle (\infty,6)\,\cup\,(6,\infty)$ Or you could write: $\displaystyle p^236>0$ $\displaystyle (p+6)(p6)>0$ Knowing the sign of the expression on the left will alternate across the 3 intervals made by the two critical values, we see that when $p=0$ the expression is negative, and so the positive intervals are: $\displaystyle (\infty,6)\,\cup\,(6,\infty)$ 
April 3rd, 2014, 09:24 AM  #4  
Newbie Joined: Apr 2014 From: The Netherlands Posts: 7 Thanks: 2  Quote:
 
April 3rd, 2014, 09:32 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
If we have: $\displaystyle p>6$ we may think of this in terms of distance: $\displaystyle \sqrt{(p0)^2}>6$ This means $p$ is a number on the real number line whose distance from 0 is greater than 6. The means $p$ is any number less than 6 or greater than 6. 
April 3rd, 2014, 09:37 AM  #6 
Newbie Joined: Apr 2014 From: The Netherlands Posts: 7 Thanks: 2  Thanks, I think that'll help me understand this 
April 3rd, 2014, 08:55 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  There wasn't any trial and error involved, I gave examples to demonstrate the places where your answer wasn't accurate. In general, the region between the roots of a quadratic have a different sign to the regions either side. This is because each root represents a crossing of the xaxis. (This is true of any order of polynomial). So being aware of where the curve goes as $x$ heads to $\infty$ and keeping track of how many roots there are either side of each value of $x$ tells you the sign of the curve at that point. This sort of thing becomes second nature with a bit of practice. 

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