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SA32

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This problem is really frustrating me. Any help is appreciated!

Answer the following questions for the function f(x) = [tex]\sin^{2}{\frac{x}{6}}[/tex] defined on the interval [-6π + 0.8, 6π/4 - 0.3].

A. f(x) is concave down on the region ____ to ____.

B. A global minimum for this function occurs at _____.

C. A local maximum for this function which is not a global maximum occurs at ____.

D. The function is increasing on ____ to ____ and on _____ to ____.

So I found the first derivative, [tex]2\sin{\frac{x}{6}}\cos{(\frac{x}{6})}\frac{1}{6}[/tex] which I thought I could simplify to [tex]\frac{1}{6}\sin{\frac{2x}{6}}[/tex]

Then taking the second derivative gives, [tex]\frac{1}{18}\cos{\frac{2x}{6}}[/tex]

Trig has never been my strong suit, and I guess the major problem I'm having here is the domain. I can't look at it and understand clearly what it means. It seems to me that there are several minima and maxima in that domain, but I don't remember sine or cosine graphs having local extrema that aren't also global extrema. I'm hazarding a guess that this means the graph isn't along the x-axis... like, it's either tilted toward the positive y values or the negative y values? Does that mean that the global minimum occurs at one of the end points? The phrasing of the question "

And for the local maximum... I think there's going to be more than one, so maybe I can just set the first derivative to zero and choose any of them.

As for concavity, I know I use the second derivative to find that, and the first derivative to find increasing/decreasing, but... I just don't know what to do in this case because I don't fully understand the domain. I tried looking at the unit circle but since the values for the domain aren't actually on it, it didn't really help.

Answer the following questions for the function f(x) = [tex]\sin^{2}{\frac{x}{6}}[/tex] defined on the interval [-6π + 0.8, 6π/4 - 0.3].

A. f(x) is concave down on the region ____ to ____.

B. A global minimum for this function occurs at _____.

C. A local maximum for this function which is not a global maximum occurs at ____.

D. The function is increasing on ____ to ____ and on _____ to ____.

So I found the first derivative, [tex]2\sin{\frac{x}{6}}\cos{(\frac{x}{6})}\frac{1}{6}[/tex] which I thought I could simplify to [tex]\frac{1}{6}\sin{\frac{2x}{6}}[/tex]

Then taking the second derivative gives, [tex]\frac{1}{18}\cos{\frac{2x}{6}}[/tex]

Trig has never been my strong suit, and I guess the major problem I'm having here is the domain. I can't look at it and understand clearly what it means. It seems to me that there are several minima and maxima in that domain, but I don't remember sine or cosine graphs having local extrema that aren't also global extrema. I'm hazarding a guess that this means the graph isn't along the x-axis... like, it's either tilted toward the positive y values or the negative y values? Does that mean that the global minimum occurs at one of the end points? The phrasing of the question "

*A*global minimum" suggests there is more than one of those, too, though. Confusing.And for the local maximum... I think there's going to be more than one, so maybe I can just set the first derivative to zero and choose any of them.

As for concavity, I know I use the second derivative to find that, and the first derivative to find increasing/decreasing, but... I just don't know what to do in this case because I don't fully understand the domain. I tried looking at the unit circle but since the values for the domain aren't actually on it, it didn't really help.

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