Identify the maximum and minimum values of the function in the interval $ [-2 \pi, 2 \pi] $. Use your understanding of transformations, not your graphing calculator.
-$$y = \frac { 1 } { 2 } \sin 2 x$$
A) Maximum value $= \frac { 1 } { 2 } ;$ minimum value $= - \frac { 1 } { 2 }$
B) Maximum value $= \frac { 1 } { 2 } ;$ minimum value $= 0$
C) Maximum value $= \frac { 1 } { 4 } ;$ minimum value $= - \frac { 1 } { 4 }$
D) Maximum value $= 2$; minimum value $= - 2$

Question

Quizplus · Accepted Answer

The function $y = \frac { 1 } { 2 } \sin 2x$ has a maximum value when $\sin 2x = 1$, which is $\frac { 1 } { 2 }$, and a minimum value when $\sin 2x = -1$, which is $-\frac { 1 } { 2 }$. The factor of $\frac { 1 } { 2 }$ scales the amplitude of the sine function, and the factor of $2$ inside the sine function affects the period but not the maximum and minimum values.

Identify the Maximum and Minimum Values of the Function in the Interval