consider the equation below. (if an answer does not exist, enter dne.) f(x) = 3 cos²(x) - 6 sin(x), 0 ≤ x ≤…

Explanation:

Step1: Find the first - derivative

First, use the chain - rule. Given $f(x)=3\cos^{2}(x)-6\sin(x)$. We know that if $u = \cos(x)$, then $\frac{d}{dx}(3\cos^{2}(x))=3\times2\cos(x)(-\sin(x))=-6\sin(x)\cos(x)$. And $\frac{d}{dx}(-6\sin(x))=-6\cos(x)$. So $f'(x)=-6\sin(x)\cos(x)-6\cos(x)=-6\cos(x)(\sin(x) + 1)$.

Step2: Find critical points

Set $f'(x)=0$. Then $-6\cos(x)(\sin(x)+1)=0$. Since $\sin(x)+1\geq0$ for all $x$, and $\sin(x)+1 = 0$ when $\sin(x)=-1$ (i.e., $x=\frac{3\pi}{2}$), and $\cos(x)=0$ when $x=\frac{\pi}{2},\frac{3\pi}{2}$.

Step3: Determine intervals of increase and decrease

Test the intervals:

• For $0\leq x<\frac{\pi}{2}$, let's take $x = \frac{\pi}{4}$. Then $f'(\frac{\pi}{4})=-6\cos(\frac{\pi}{4})(\sin(\frac{\pi}{4}) + 1)<0$, so $f(x)$ is decreasing on $[0,\frac{\pi}{2})$.
• For $\frac{\pi}{2}<x<\frac{3\pi}{2}$, let's take $x=\pi$. Then $f'(\pi)=-6\cos(\pi)(\sin(\pi)+1)=6>0$, so $f(x)$ is increasing on $(\frac{\pi}{2},\frac{3\pi}{2})$.
• For $\frac{3\pi}{2}<x\leq2\pi$, let's take $x=\frac{7\pi}{4}$. Then $f'(\frac{7\pi}{4})=-6\cos(\frac{7\pi}{4})(\sin(\frac{7\pi}{4})+1)<0$, so $f(x)$ is decreasing on $(\frac{3\pi}{2},2\pi]$.

Step4: Find local minima and maxima

Evaluate $f(x)$ at critical points and endpoints:

• $f(0)=3\cos^{2}(0)-6\sin(0)=3$.
• $f(\frac{\pi}{2})=3\cos^{2}(\frac{\pi}{2})-6\sin(\frac{\pi}{2})=-6$.
• $f(\frac{3\pi}{2})=3\cos^{2}(\frac{3\pi}{2})-6\sin(\frac{3\pi}{2})=6$.
• $f(2\pi)=3\cos^{2}(2\pi)-6\sin(2\pi)=3$. So the local minimum value is $-6$ and the local maximum value is $6$.

Step5: Find the second - derivative

$f'(x)=-6\sin(x)\cos(x)-6\cos(x)=-3\sin(2x)-6\cos(x)$. Then $f''(x)=-6\cos(2x)+6\sin(x)$. Using the double - angle formula $\cos(2x)=1 - 2\sin^{2}(x)$, we have $f''(x)=-6(1 - 2\sin^{2}(x))+6\sin(x)=12\sin^{2}(x)+6\sin(x)-6 = 6(2\sin^{2}(x)+\sin(x)-1)=6(2\sin(x)-1)(\sin(x)+1)$.

Step6: Find inflection points

Set $f''(x)=0$. Then $6(2\sin(x)-1)(\sin(x)+1)=0$. So $\sin(x)=\frac{1}{2}$ or $\sin(x)=-1$. When $\sin(x)=\frac{1}{2}$, $x=\frac{\pi}{6},\frac{5\pi}{6}$; when $\sin(x)=-1$, $x = \frac{3\pi}{2}$.

• $f(\frac{\pi}{6})=3\cos^{2}(\frac{\pi}{6})-6\sin(\frac{\pi}{6})=3\times\frac{3}{4}-6\times\frac{1}{2}=-\frac{3}{4}$.
• $f(\frac{5\pi}{6})=3\cos^{2}(\frac{5\pi}{6})-6\sin(\frac{5\pi}{6})=3\times\frac{3}{4}-6\times\frac{1}{2}=-\frac{3}{4}$.
• $f(\frac{3\pi}{2})=6$. The inflection points are $(\frac{\pi}{6},-\frac{3}{4}),(\frac{5\pi}{6},-\frac{3}{4}),(\frac{3\pi}{2},6)$.

Step7: Determine concavity intervals

Test the intervals:

• For $0\leq x<\frac{\pi}{6}$, let's take $x=\frac{\pi}{12}$. Then $f''(\frac{\pi}{12})=6(2\sin(\frac{\pi}{12})-1)(\sin(\frac{\pi}{12})+1)<0$, so $f(x)$ is concave down on $[0,\frac{\pi}{6})$.
• For $\frac{\pi}{6}<x<\frac{5\pi}{6}$, let's take $x=\frac{\pi}{2}$. Then $f''(\frac{\pi}{2})=6(2\sin(\frac{\pi}{2})-1)(\sin(\frac{\pi}{2})+1)=12>0$, so $f(x)$ is concave up on $(\frac{\pi}{6},\frac{5\pi}{6})$.
• For $\frac{5\pi}{6}<x<\frac{3\pi}{2}$, let's take $x=\pi$. Then $f''(\pi)=6(2\sin(\pi)-1)(\sin(\pi)+1)= - 6<0$, so $f(x)$ is concave down on $(\frac{5\pi}{6},\frac{3\pi}{2})$.
• For $\frac{3\pi}{2}<x\leq2\pi$, let's take $x=\frac{7\pi}{4}$. Then $f''(\frac{7\pi}{4})=6(2\sin(\frac{7\pi}{4})-1)(\sin(\frac{7\pi}{4})+1)<0$, so $f(x)$ is concave down on $(\frac{3\pi}{2},2\pi]$.

Answer:

(a) Increasing: $(\frac{\pi}{2},\frac{3\pi}{2})$; Decreasing: $[0,\frac{\pi}{2})\cup(\frac{3\pi}{2},2\pi]$ (b) Local minimum value: $-6$; Local maximum value: $6$ (c) Inflection points: $(\frac{\pi}{6},-\frac{3}{4}),(\frac{5\pi}{6},-\frac{3}{4}),(\frac{3\pi}{2},6)$; Concave up: $(\frac{\pi}{6},\frac{5\pi}{6})$; Concave down: $[0,\frac{\pi}{6})\cup(\frac{5\pi}{6},\frac{3\pi}{2})\cup(\frac{3\pi}{2},2\pi]$