### Video Transcript

The range of the function π of π equals π sin two π is the closed interval from negative five to five. Find the value of π, where π is greater than zero.

So weβve been given the range of this function π of π and asked to use this information to determine the value of the unknown π. We recall that the sine function is periodic with a period of two π and its graph oscillates between negative one and positive one. In other words, the range of the sine function is the closed interval from negative one to positive one.

Now, we arenβt just working with a sine function in this question. Weβre working with a transformation of this. π of π is equal to π multiplied by sin of two π. Two changes have been made. Firstly, the variable π has been replaced with two π. This corresponds to a horizontal dilation or stretch of the function with a scale factor of one-half. As this transformation has a horizontal effect, it has no effect on the functionβs range.

The other transformation that has been applied is the function has been multiplied by the constant π. This corresponds to a vertical stretch of the function by a scale factor π. When we stretch a function vertically by a scale factor of π, this has the effect of multiplying the values in its range by π. So, if the range of the sine function is the closed interval from negative one to positive one, the range of the function π of π equals π sin of two π is the closed interval from negative π to π. And weβre told that this must be equal to the closed interval from negative five to five.

Now, itβs really important to note that we were told that π is positive. This means that when we multiply the sine function by π, this only has the effect of stretching the function by a factor of π. If π were negative, the function would be stretched by a factor equal to the absolute value of π and would also be reflected in the π₯-axis. This would mean that the smallest value in the range would be mapped to the maximum value in the new range, and vice versa. So the endpoints would effectively swap around. But as π is positive, this isnβt the case here. So we can just compare the lower endpoint of the two intervals and the upper endpoint of the two intervals. And in doing so, we find that the value of π is five.