Why have these endpoints been excluded?

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Endpoints 0 and 24 are excluded from the derivative of the piecewise function because the derivative requires the function to be defined in a neighborhood around those points. Although f(h) includes 0 and 24, the continuity of the derivative is crucial, and the derivative is not defined at h = 9/4 due to discontinuity. The discussion emphasizes that while the endpoints' inclusion may seem relevant, it is not critical for determining the function's minimum and maximum. The focus remains on where the derivative is zero or undefined to identify these extrema. Understanding these exclusions clarifies the behavior of the derivative at specific points.
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Homework Statement
The problem asks for the derivative of a piecewise function. I don't know why it has excluded the endpoints from the correct answer.
Relevant Equations
f(h) = 4h+10 0<=h<=9/4
f(h) = 0.5sin[2pi/5(h-9/4)]+19 9/4<h<=24
Hi everyone

Could someone please confirm that these solutions are correct? They exclude the endpoints 0 and 24, but I don't understand why, since f(h) includes 0 and 24.

image_2022-05-23_165844737.png


I understand why 9/4 is excluded, as the derivative is not a continuous function.

Thanks
 
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Darkmisc said:
Homework Statement:: The problem asks for the derivative of a piecewise function. I don't know why it has excluded the endpoints from the correct answer.
Generally the derivative of a function ##f## at the point ##a## requires the function to be defined in some neighbourhood of ##a## (i.e. open interval containing ##a##). That said, it's not very important in this case whether we consider ##f'## to be defined at ##0## and ##24## or not.

Darkmisc said:
I understand why 9/4 is excluded, as the derivative is not a continuous function.
Technically the derivative is not defined for ##h = \frac 9 4##.
 
PeroK said:
That said, it's not very important in this case whether we consider ##f'## to be defined at ##0## and ##24## or not.
The minimum and maximum of a function occur where its derivative is 0 or undefined.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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