For the function y= -2+5sin (π/12 (x-2)), what is the minimum value?

The value of a sine function is between -1 and 1.
-1 ≤ sin [π/12 (x - 2)] ≤ 1
-5 ≤ 5 sin [π/12 (x - 2)] ≤ 5
-2 + (-5) ≤ -2 + 5 sin [π/12 (x - 2)] ≤ -2 + 5
-7 ≤ -2 + 5 sin [π/12 (x - 2)] ≤ 3
-7 ≤ y ≤ 3

Hence, the minimum value of y = -7

min(y) = -2 +5*min(sin( )) = -2 +5*(-1) = -7

The minimum value of the function is y = -7.
It is found at x = 24n-4, where n is any integer.

Differentiate and equate to zero.
Us the Chain Rule
dy/dx = dy/du X du/dx = 0

y = -2 + 5Sin((pi/12)x - pi/6)
Let u = (pi/12) - pi/6

Hence
y = -2 + 5 SinU
dy/du = 5CosU

&
du/dx= pi/12

Hence
dy/dx = 5Cos(U) X pi/12
dy/dx = 5Cos((pi/12)x - pi/6) X pi/12 = 0
(5pi/12) Cos ((pi/12)x - pi/6) = 0
Cos((pi/12)x - pi/6) = 0
(pi/12)x - pi/6 = Cos^-1(0) = 1
(pi/12)x = 1 + pi/6
x = (1 + pi/6) / pi/12)
x = [(6 + pi)/ 6] / pi/12
x = (6 + pi) / 6 X 12 / pi
x = 2(6 + pi) / pi
x = (12 + 2pi) / pi radians.