How to tell if a function's derivative is always positive?

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The discussion centers on determining whether the derivative of the function \(\frac{dP}{dt} = P(a - bP)\) is always positive between its critical points \(P = 0\) and \(P = \frac{a}{b}\). It is established that by selecting an arbitrary point, such as \(P = \frac{a}{2b}\), and evaluating the derivative at this point, one can confirm that \(P'(\frac{a}{2b}) = \frac{a^2}{4b} > 0\). This indicates that the derivative remains positive between the critical points, given that \(b > 0\) ensures the parabola is concave down.

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find_the_fun
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In class we were given an example where [math]\frac{dP}{dt}=P(a-bP)[/math]. We found the critical points to be P=0 and P=a/b. We wanted to know if the derivative is always positive or negative between the two critical points. The prof said you could pick an arbitrary point between the two, such as [math]\frac{a}{2b}[/math] and plug that into the derivative and check to see if it's greater than 0. So [math]P'(\frac{2}{2b})=\frac{a^2}{4b} > 0[/math].

I'm really confused and don't understand the reasoning, can someone fill in the gaps?
 
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find_the_fun said:
In class we were given an example where [math]\frac{dP}{dt}=P(a-bP)[/math]. We found the critical points to be P=0 and P=a/b. We wanted to know if the derivative is always positive or negative between the two critical points. The prof said you could pick an arbitrary point between the two, such as [math]\frac{a}{2b}[/math] and plug that into the derivative and check to see if it's greater than 0. So [math]P'(\frac{2}{2b})=\frac{a^2}{4b} > 0[/math].

I'm really confused and don't understand the reasoning, can someone fill in the gaps?

Hi find_the_fun,

$$P(a-bP)$$ is a parabola.
Its top is between $P=0$ and $P=a/b$, and actually at $P=\frac{a}{2b}$.
Either way, the sign of the derivative, which is equal to $P(a-bP)$, between those 2 points is either always positive, or always negative.
Testing it at some point between those boundaries will tell us which it is.

Alternatively, we can see that if $b >0$, the parabola is upside-down, meaning it is always positive between the 2 points.
 
I like Serena said:
the sign of the derivative...between those 2 points is either always positive, or always negative.

How do you know this? Is it because we know [math]\frac{dP}{dt}[/math] is not 0 between the two points, therefore it can't change sign?
 
Last edited:
find_the_fun said:
How do you know this? Is it because we know [math]\frac{dP}{dt}[/math] is not 0 between the two points, therefore it can't change sign?

Yes.
(With the additional information that it is defined and continuous between those 2 points.)
 

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