Why do we include the opposite value in the circle unit of a critical number if it falls in the domain?

Question

f(t) = 9t + 9cot(t/2). Find the absolute minimum and maximum values in the domain [π/4, 7π/4]

the derivative of (df/dt)f(t) = 9+(-9csc^2(t/2)*1/2).

When we set the derivative of f(t)=0, then:

t = π/2

Now I understand 3π/2 falls in the domain however, why do we include it as a critical value of x? It cannot simply be because it runs parallel to the critical value of π/2 otherwise, what is the import or significance of it being parallel?

Answer 1

Here's a plot of your function (blue) and its derivative (orange):

Because the derivative is always positive, the minimum of your function is at the left (at $\pi/4$) and the maximum is at the right (at $7 \pi/4$).

$f(t) = 9 t + \cot \left( {t \over 2} \right)$

and

${\partial \over \partial t} f(t) = 9 - {1 \over 2} \csc^2 \left( {t \over 2} \right)$