Question - Finding Absolute Maximum and Minimum of a Trigonometric Function
Solution:
The question asks for the absolute maximum and minimum of the function $$ f(x) = 3\cos^2\left(\frac{x}{2}\right) $$ over the interval $$\left[\frac{\pi}{4}, \pi\right]$$.To solve this:1. Compute the derivative of the function to find the critical points where the derivative is zero or undefined. Critical points can also occur at the endpoints of the interval.2. Evaluate the function at the critical points and at the endpoints of the interval to determine the absolute maximum and minimum.Let's find the derivative $$f'(x)$$:$$ f(x) = 3\cos^2\left(\frac{x}{2}\right) $$Using the chain rule,$$ f'(x) = 3 \cdot 2\cos\left(\frac{x}{2}\right)(-\sin\left(\frac{x}{2}\right)) \left(\frac{1}{2}\right) = -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) $$Setting the derivative equal to zero,$$ -3\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = 0 $$This equation is satisfied when either $$ \sin\left(\frac{x}{2}\right) = 0 $$ or $$ \cos\left(\frac{x}{2}\right) = 0 $$.$$ \sin\left(\frac{x}{2}\right) = 0 $$ at $$ x = 0, 2\pi, 4\pi, \ldots $$ but within $$\left[\frac{\pi}{4}, \pi\right]$$, this doesn't give any solution.$$ \cos\left(\frac{x}{2}\right) = 0 $$ at $$ x = \pi, 3\pi, \ldots $$, and only $$ x = \pi $$ falls within our interval.The critical points are just at the endpoints of the interval since there are no other points in $$\left[\frac{\pi}{4}, \pi\right]$$ where the derivative is zero or undefined. Now, evaluate $$f(x)$$ at $$ x = \frac{\pi}{4} $$ and $$ x = \pi $$:$$ f\left(\frac{\pi}{4}\right) = 3\cos^2\left(\frac{\pi}{8}\right) $$ and $$f(\pi) = 3\cos^2\left(\frac{\pi}{2}\right)$$.Note that $$ \cos\left(\frac{\pi}{2}\right) = 0 $$, so $$ f(\pi) = 0 $$.To find $$f\left(\frac{\pi}{4}\right)$$, we compute the cosine of $$ \pi/8 $$. This value is not one of the standard angles, so we can't find an exact value easily without using a calculator. However, since $$0 < \frac{\pi}{8} < \frac{\pi}{2}$$, $$ \cos\left(\frac{\pi}{8}\right) $$ is positive and less than 1, so $$ 3\cos^2\left(\frac{\pi}{8}\right) $$ is also positive and less than 3.Therefore, the absolute maximum of $$ f(x) $$ is at $$ x = \frac{\pi}{4} $$ and is greater than 0, and the absolute minimum of $$ f(x) $$ is 0 at $$ x = \pi $$. Based only on the information available, you can't give an exact value for the absolute maximum without a calculator, but you can state where it occurs and that the absolute minimum is 0 at $$ x = \pi $$.
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