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

[find_the_fun]

In class we were given an example where \(\displaystyle \frac{dP}{dt}=P(a-bP)\). 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 \(\displaystyle \frac{a}{2b}\) and plug that into the derivative and check to see if it's greater than 0. So \(\displaystyle P'(\frac{2}{2b})=\frac{a^2}{4b} > 0\).

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

 

[I like Serena]

In class we were given an example where \(\displaystyle \frac{dP}{dt}=P(a-bP)\). 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 \(\displaystyle \frac{a}{2b}\) and plug that into the derivative and check to see if it's greater than 0. So \(\displaystyle P'(\frac{2}{2b})=\frac{a^2}{4b} > 0\).

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

Hi find_the_fun,

\(\displaystyle P(a-bP)\) 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.

 

[find_the_fun]

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 \(\displaystyle \frac{dP}{dt}\) is not 0 between the two points, therefore it can't change sign?

 

[I like Serena]

How do you know this? Is it because we know \(\displaystyle \frac{dP}{dt}\) 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.)

 

[blue_raver22]

I can provide an explanation for the method suggested by your professor to determine if the derivative of a function is always positive. In the example given, the derivative of the function is represented as dP/dt, which is equal to P(a-bP). The critical points of this function are P=0 and P=a/b.

To determine if the derivative is always positive or negative between these two critical points, your professor suggested picking an arbitrary point between them, such as a/2b, and plugging it into the derivative. This means substituting P=a/2b into the derivative equation, which gives us P'(a/2b) = (a/2b)(a-b(a/2b)). Simplifying this gives us P'(a/2b) = (a^2)/4b.

The reasoning behind this method is that if the derivative at this arbitrary point, P'(a/2b), is greater than 0, then it means the derivative must be positive between the two critical points. This is because the derivative is a measure of the rate of change of the function at a specific point, and if the derivative is positive, it means the function is increasing at that point.

On the other hand, if the derivative at this arbitrary point is negative, it means the function is decreasing at that point, and therefore the derivative must be negative between the two critical points. In this way, checking the derivative at an arbitrary point allows us to determine the overall trend of the derivative between the critical points.

I hope this explanation helps to clarify the method suggested by your professor. Overall, it is a valid and commonly used approach in mathematics and science to determine the behavior of a function's derivative between critical points.