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April 3rd, 2014, 07:58 AM   #1
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Smile 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
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April 3rd, 2014, 08:49 AM   #2
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Quote:
Originally Posted by jyrdo View Post
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 View Post
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 $|p| \gt 6$

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Originally Posted by jyrdo View Post
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.
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April 3rd, 2014, 08:56 AM   #3
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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:

$\displaystyle (-\infty,-6)\,\cup\,(6,\infty)$
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April 3rd, 2014, 09:24 AM   #4
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Quote:
Originally Posted by v8archie View Post
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.
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April 3rd, 2014, 09:32 AM   #5
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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
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April 3rd, 2014, 09:37 AM   #6
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Quote:
Originally Posted by MarkFL View Post
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
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April 3rd, 2014, 08:55 PM   #7
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Originally Posted by jyrdo View Post
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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