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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
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Quote:
 Originally Posted by jyrdo So, we know that the discriminant must be greater than zero.
This guarantees that the function has real roots. To have a negative minimum also requires that a>0.

Quote:
 Originally Posted by jyrdo p² > 36
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 $|p| \gt 6$

Quote:
 Originally Posted by jyrdo 4p (p - 3) > 0
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 $4p \lt 0, p-3 \lt 0$ 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: $\displaystyle p^2-36>0$ $\displaystyle (p+6)(p-6)>0$ 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
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Quote:
 Originally Posted by v8archie 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 $4p \lt 0, p-3 \lt 0$ 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.
Do you have to do this by trial and error everytime? I know this is pretty basic but I'm still having a hard time understanding it. I get that |p| = 6, means that p is either 6 or -6, but the changing of the signs confuses me.

 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: $\displaystyle \sqrt{(p-0)^2}>6$ 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. Thanks from jyrdo
April 3rd, 2014, 09:37 AM   #6
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Quote:
 Originally Posted by MarkFL If we have: $\displaystyle |p|>6$ we may think of this in terms of distance: $\displaystyle \sqrt{(p-0)^2}>6$ 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.
Thanks, I think that'll help me understand this

April 3rd, 2014, 08:55 PM   #7
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Quote:
 Originally Posted by jyrdo Do you have to do this by trial and error everytime?
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 x-axis. (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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