consider the following. (if an answer does not exist, enter dne.) f(x)=3sin(x)+3cos(x), 0≤x≤2π (a) find the…

Explanation:

Step1: Find the first - derivative

Given $f(x)=3\sin(x)+3\cos(x)$, using the derivative rules $\frac{d}{dx}\sin(x)=\cos(x)$ and $\frac{d}{dx}\cos(x)=-\sin(x)$, we have $f^\prime(x)=3\cos(x)-3\sin(x)$.

Step2: Find critical points

Set $f^\prime(x) = 0$, so $3\cos(x)-3\sin(x)=0$. Then $\cos(x)=\sin(x)$, which gives $\tan(x) = 1$. In the interval $0\leq x\leq2\pi$, $x=\frac{\pi}{4},\frac{5\pi}{4}$.

Step3: Determine increasing and decreasing intervals

Test intervals:

• For $0\leq x<\frac{\pi}{4}$, let's take $x = 0$. Then $f^\prime(0)=3\cos(0)-3\sin(0)=3>0$, so $f(x)$ is increasing on $[0,\frac{\pi}{4})$.
• For $\frac{\pi}{4}<x<\frac{5\pi}{4}$, let's take $x=\pi$. Then $f^\prime(\pi)=3\cos(\pi)-3\sin(\pi)=- 3<0$, so $f(x)$ is decreasing on $(\frac{\pi}{4},\frac{5\pi}{4})$.
• For $\frac{5\pi}{4}<x\leq2\pi$, let's take $x=\frac{7\pi}{4}$. Then $f^\prime(\frac{7\pi}{4})=3\cos(\frac{7\pi}{4})-3\sin(\frac{7\pi}{4})=3\times\frac{\sqrt{2}}{2}-3\times(-\frac{\sqrt{2}}{2}) = 3\sqrt{2}>0$, so $f(x)$ is increasing on $(\frac{5\pi}{4},2\pi]$.

Step4: Find local - extrema

Since $f(x)$ changes from increasing to decreasing at $x = \frac{\pi}{4}$, $f(\frac{\pi}{4})=3\sin(\frac{\pi}{4})+3\cos(\frac{\pi}{4})=3\sqrt{2}$ is a local maximum. Since $f(x)$ changes from decreasing to increasing at $x=\frac{5\pi}{4}$, $f(\frac{5\pi}{4})=3\sin(\frac{5\pi}{4})+3\cos(\frac{5\pi}{4})=-3\sqrt{2}$ is a local minimum.

Step5: Find the second - derivative

$f^\prime(x)=3\cos(x)-3\sin(x)$, then $f^{\prime\prime}(x)=-3\sin(x)-3\cos(x)$.

Step6: Find inflection points

Set $f^{\prime\prime}(x)=0$, so $-3\sin(x)-3\cos(x)=0$, which gives $\tan(x)=-1$. In the interval $0\leq x\leq2\pi$, $x=\frac{3\pi}{4},\frac{7\pi}{4}$. $f(\frac{3\pi}{4})=3\sin(\frac{3\pi}{4})+3\cos(\frac{3\pi}{4}) = 0$ and $f(\frac{7\pi}{4})=3\sin(\frac{7\pi}{4})+3\cos(\frac{7\pi}{4})=0$. So the inflection points are $(\frac{3\pi}{4},0)$ and $(\frac{7\pi}{4},0)$.

Step7: Determine concavity intervals

Test intervals:

• For $0\leq x<\frac{3\pi}{4}$, let's take $x=\frac{\pi}{2}$. Then $f^{\prime\prime}(\frac{\pi}{2})=-3\sin(\frac{\pi}{2})-3\cos(\frac{\pi}{2})=-3<0$, so $f(x)$ is concave down on $[0,\frac{3\pi}{4})$.
• For $\frac{3\pi}{4}<x<\frac{7\pi}{4}$, let's take $x=\pi$. Then $f^{\prime\prime}(\pi)=-3\sin(\pi)-3\cos(\pi)=3>0$, so $f(x)$ is concave up on $(\frac{3\pi}{4},\frac{7\pi}{4})$.
• For $\frac{7\pi}{4}<x\leq2\pi$, let's take $x=\frac{15\pi}{8}$. Then $f^{\prime\prime}(\frac{15\pi}{8})=-3\sin(\frac{15\pi}{8})-3\cos(\frac{15\pi}{8})=-3\times(-\frac{\sqrt{2 + \sqrt{2}}}{2})-3\times\frac{\sqrt{2-\sqrt{2}}}{2}<0$, so $f(x)$ is concave down on $(\frac{7\pi}{4},2\pi]$.

Answer:

(a) Increasing: $[0,\frac{\pi}{4})\cup(\frac{5\pi}{4},2\pi]$; Decreasing: $(\frac{\pi}{4},\frac{5\pi}{4})$ (b) Local minimum value: $-3\sqrt{2}$; Local maximum value: $3\sqrt{2}$ (c) Inflection points: $(\frac{3\pi}{4},0),(\frac{7\pi}{4},0)$; Concave up: $(\frac{3\pi}{4},\frac{7\pi}{4})$; Concave down: $[0,\frac{3\pi}{4})\cup(\frac{7\pi}{4},2\pi]$