Question

sketch the graph of the following function and determine whether the function has any absolute extreme values on its domain. explain how your answer is consistent with the extreme value theorem.
y = 5 sin x, 0 < x < 2π
c. the function has an absolute maximum value at x - but does not have an absolute minimum value on its domain.
d. the function does not have any absolute extreme values on its domain.
explain the results in terms of the extreme value theorem.
a. since the function f is continuous on a closed interval, it attains both an absolute maximum value and an absolute minimum value on its domain.
b. since the function f is continuous on an open interval, it may or may not have any absolute extreme values on its domain.
c. since the function f is not continuous on a closed interval, it may or may not have any absolute extreme values on its domain.
d. since the function f is not continuous on an open interval, it does not attain any absolute extreme values on its domain.

Explanation:

Step1: Recall properties of sine function

The general form of a sine - function is $y = A\sin(x)$, where $A$ is the amplitude. For $y = 5\sin(x)$, the amplitude $A = 5$, and the period of $y=\sin(x)$ is $2\pi$, so the period of $y = 5\sin(x)$ is also $2\pi$.

Step2: Analyze extreme values

The function $y = 5\sin(x)$ has a maximum value of $y = 5$ when $\sin(x)=1$ (i.e., $x=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$) and a minimum value of $y=- 5$ when $\sin(x)=-1$ (i.e., $x=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$). In the open interval $(0,2\pi)$, the maximum value of $y = 5\sin(x)$ is $y = 5$ at $x=\frac{\pi}{2}$ and the minimum value is $y=-5$ at $x = \frac{3\pi}{2}$.

Step3: Apply extreme - value theorem

The extreme - value theorem states that if a function $y = f(x)$ is continuous on a closed interval $[a,b]$, then $f(x)$ attains both an absolute maximum and an absolute minimum on $[a,b]$. The function $y = 5\sin(x)$ is continuous on the open interval $(0,2\pi)$. For an open interval, a function may or may not have absolute extreme values. In this case, $y = 5\sin(x)$ on $(0,2\pi)$ has extreme values because of the nature of the sine function, but in general, for a continuous function on an open interval, extreme values are not guaranteed.

Answer:

The function has an absolute maximum value of $5$ at $x=\frac{\pi}{2}$ and an absolute minimum value of $- 5$ at $x=\frac{3\pi}{2}$ in the domain $(0,2\pi)$. For the explanation part: B. Since the function $f$ is continuous on an open interval, it may or may not have any absolute extreme values on its domain.