the polar function $r = f(\\theta)$, where $f(\\theta)=1 + 2\\cos\\theta$, is graphed in the polar…

Explanation:

Step1: Recall the distance in polar - coordinates

In polar coordinates, the distance between a point $(r,\theta)$ and the origin is given by $r$. Here, $r = f(\theta)=1 + 2\cos\theta$.

Step2: Find the derivative of $r$ with respect to $\theta$

Differentiate $r = 1+2\cos\theta$ with respect to $\theta$. Using the derivative formula $\frac{d}{d\theta}\cos\theta=-\sin\theta$, we get $r'=- 2\sin\theta$.

Step3: Determine when $r$ is decreasing

A function $y = r(\theta)$ is decreasing when $r'(\theta)<0$. So we set $-2\sin\theta<0$, which simplifies to $\sin\theta>0$. We know that $\sin\theta>0$ for $0 <\theta<\pi$ and $\sin\theta < 0$ for $\pi<\theta<2\pi$. We also need to find the critical - points of $r = 1 + 2\cos\theta$ by setting $r'=-2\sin\theta = 0$. The solutions are $\theta = 0,\pi,2\pi$. Now, we can analyze the sign of $r'$ on sub - intervals of $[0,2\pi]$. Let's consider the intervals:

• For $\theta\in(0,\pi)$, $\sin\theta>0$, so $r'=-2\sin\theta<0$ and $r$ is decreasing.
• For $\theta\in(\pi,2\pi)$, $\sin\theta<0$, so $r'=-2\sin\theta>0$ and $r$ is increasing. We can convert the radians to decimal approximations. $\pi\approx3.142$. We know that $r = 1 + 2\cos\theta$ is decreasing when $\sin\theta>0$. The interval where $\sin\theta>0$ in the given options and in the domain $0\leq\theta\leq2\pi$ is when $\theta\in(2.094,3.142)$ and $\theta\in(4.189,6.283)$ (since $2.094\approx\frac{2\pi}{3}$, $3.142\approx\pi$, $4.189\approx\frac{4\pi}{3}$, $6.283\approx2\pi$).

Answer:

D. $(2.094,3.142)$ and $(4.189,6.283)$