Question

a. Find the critical points of the following functions on the given interval. b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither. c. Find the absolute maximum and minimum values on the given interval when they exist. $$f(\theta)=2 \sin \theta+\cos \theta \text { on }[-2 \pi, 2 \pi]$$

Short answer

Solution: The absolute maximum value of the function occurs at $$\theta=\frac{\pi}{3}$$, with an absolute maximum value of $$f(\frac{\pi}{3})=\sqrt{3}+\frac{1}{2}$$, and the absolute minimum value of the function occurs at $$\theta=-\frac{\pi}{2}$$, with an absolute minimum value of $$f(-\frac{\pi}{2})=-2$$.

Step-by-step solution

Compute the first derivative of the function

To find the critical points of the function, compute the first derivative of $$f(\theta)=2 \sin \theta+\cos \theta$$ with respect to $$\theta$$: $$f'(\theta)=\frac{d}{d \theta} (2 \sin \theta+\cos \theta)=2\cos\theta-\sin\theta$$

Find the critical points

To find the critical points, set the first derivative equal to zero and solve for $$\theta$$: $$f'(\theta)=0 \Rightarrow 2\cos\theta-\sin\theta=0$$ This trigonometric equation can be solved using a substitution, such as: $$u=\cos\theta \Rightarrow \sin\theta=u\cos^{-1}(u)$$ Substitute this into the equation: $$2u-u\cos^{-1}(u)=0 \Rightarrow u(2-\cos^{-1}(u))=0$$ This gives us two possible cases: 1. $$u=0 \Rightarrow \cos\theta=0 \Rightarrow \theta=\pm\frac{\pi}{2}$$ or, 2. $$\cos^{-1}(u)=2 \Rightarrow \cos\theta=\frac{1}{2} \Rightarrow \theta=\pm\frac{\pi}{3}$$ So, the critical points are at $$\theta=\pm\frac{\pi}{2}$$ and $$\pm\frac{\pi}{3}$$.

Determine the nature of the critical points

Now that we have the critical points, we will use the second derivative test to determine whether these points correspond to local maxima, local minima, or neither. First, compute the second derivative: $$f''(\theta)=-2\sin\theta-\cos\theta$$ Now, test the critical points with the second derivative: 1. $$f''(\pm\frac{\pi}{2})=-2\sin(\pm\frac{\pi}{2})-\cos(\pm\frac{\pi}{2})=-2(\pm 1)-0=\mp 2$$ Since $$f''(\frac{\pi}{2})<0$$, there is a local maximum at $$\theta=\frac{\pi}{2}$$. Since $$f''(-\frac{\pi}{2})>0$$, there is a local minimum at $$\theta=-\frac{\pi}{2}$$. 2. $$f''(\pm\frac{\pi}{3})=-2\sin(\pm\frac{\pi}{3})-\cos(\pm\frac{\pi}{3})=-2(\pm\frac{\sqrt{3}}{2})-\frac{1}{2}=\mp\sqrt{3}-\frac{1}{2}$$ Since $$f''(\frac{\pi}{3})<0$$, there is a local maximum at $$\theta=\frac{\pi}{3}$$. Since $$f''(-\frac{\pi}{3})>0$$, there is a local minimum at $$\theta=-\frac{\pi}{3}$$.

Find the absolute maximum and minimum values on the given interval

We have established that the critical points are at $$\theta=\pm\frac{\pi}{2}$$ and $$\pm\frac{\pi}{3}$$. To find the absolute maximum and minimum values of the function on the given interval $$[-2 \pi, 2 \pi]$$, we need to compute the function values at these points and the endpoints of the interval: $$f(\pm\frac{\pi}{2})=2 \sin (\pm\frac{\pi}{2})+\cos (\pm\frac{\pi}{2})=2(\pm 1)+0=\pm 2$$ $$f(\pm\frac{\pi}{3})=2 \sin (\pm\frac{\pi}{3})+\cos (\pm\frac{\pi}{3})=2(\pm\frac{\sqrt{3}}{2})+\frac{1}{2}=\pm\sqrt{3}+\frac{1}{2}$$ $$f(-2\pi)=2 \sin (-2\pi)+\cos (-2\pi)=2(0)+1=1$$ $$f(2\pi)=2 \sin (2\pi)+\cos (2\pi)=2(0)+1=1$$ Since $$f(\frac{\pi}{3})>f(\frac{\pi}{2})>f(2\pi)$$, the absolute maximum value of the function is at $$\theta=\frac{\pi}{3}$$, with an absolute maximum value of $$f(\frac{\pi}{3})=\sqrt{3}+\frac{1}{2}$$. Similarly, the absolute minimum value of the function is at $$\theta=-\frac{\pi}{2}$$, with an absolute minimum value of $$f(-\frac{\pi}{2})=-2$$.

Key concepts

First Derivative Test

Understanding the first derivative test is key to analyzing the behavior of functions. When we compute the derivative of a function and find where it equals zero, we've identified critical points where the function’s slope is flat, indicating potential local maxima or minima. To apply the first derivative test, we check the sign of the derivative just before and after each critical point. If it changes from negative to positive, we've found a local minimum. Conversely, if the sign changes from positive to negative, it's a local maximum. This test directly helps in mapping the nature of the graph and provides insights into its curvature at specific angles or values. For trigonometric functions like
\( f(\theta)=2 \sin \theta + \cos \theta \),
we differentiate with respect to \( \theta \) to find potential turning points for further analysis.

Second Derivative Test

The second derivative test is an excellent tool for determining concavity and inflection points. Where the first derivative test identifies critical points, the second derivative reveals whether these points are concave up, indicating a local minimum, or concave down, representing a local maximum. For the function
\( f(\theta) = 2 \sin \theta + \cos \theta \),
once we find the critical points, we compute the second derivative
\( f''(\theta) \).
By substituting the values of \( \theta \) at the critical points into the second derivative and examining the result, we can confirm the nature of each critical point found with the first derivative test.

Absolute Maximum and Minimum Values

When it comes to identifying the absolute maximum and minimum values of a function on a closed interval, we need to evaluate the function at critical points as well as the endpoints of the interval. Absolute extrema are the highest and lowest points of the function over this range. For example, in the function
\( f(\theta) = 2 \sin \theta + \cos \theta \),
we determine the values of \( f(\theta) \) at the critical points \( \theta = \pm\frac{\pi}{2} \) and \( \theta = \pm\frac{\pi}{3} \) alongside the interval endpoints \( \theta = -2\pi \) and \( \theta = 2\pi \). The highest and lowest of these values give us the absolute maximum and minimum values, respectively.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is a fundamental process in calculus. For instance, to differentiate
\( f(\theta)=2 \sin \theta + \cos \theta \),
we apply the derivatives of basic trigonometric functions: the derivative of \( \sin \theta \) is \( \cos \theta \), and the derivative of \( \cos \theta \) is \( -\sin \theta \). Consequently, the derivative of our function becomes
\( f'(\theta)=2\cos\theta - \sin\theta \).
Understanding these principles allows us to determine the rate of change at any point on a curve defined by trigonometric expressions, which is essential for finding critical points and analyzing the function's behavior.