consider the equation below. (if an answer does not exist, enter dne.) f(x) = 5 cos²(x) - 10 sin(x), 0 ≤ x ≤…

Explanation:

Step1: Find the first - derivative

First, use the chain - rule and basic derivative formulas. Given $f(x)=5\cos^{2}(x)-10\sin(x)$. Let $u = \cos(x)$, then $\cos^{2}(x)=u^{2}$. The derivative of $u^{2}$ with respect to $x$ is $2u\cdot(-\sin(x))=-2\cos(x)\sin(x)$ by the chain - rule. The derivative of $- 10\sin(x)$ is $-10\cos(x)$. So, $f^\prime(x)=-10\cos(x)\sin(x)-10\cos(x)=-10\cos(x)(\sin(x) + 1)$.

Step2: Find critical points

Set $f^\prime(x)=0$. Then $-10\cos(x)(\sin(x)+1)=0$. 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 increasing and decreasing intervals

Test the intervals $[0,\frac{\pi}{2})$, $(\frac{\pi}{2},\frac{3\pi}{2})$, and $(\frac{3\pi}{2},2\pi]$.

• For $x\in[0,\frac{\pi}{2})$, let's take $x = \frac{\pi}{4}$. Then $f^\prime(\frac{\pi}{4})=-10\cos(\frac{\pi}{4})(\sin(\frac{\pi}{4}) + 1)<0$, so $f(x)$ is decreasing on $[0,\frac{\pi}{2})$.
• For $x\in(\frac{\pi}{2},\frac{3\pi}{2})$, let's take $x=\pi$. Then $f^\prime(\pi)=-10\cos(\pi)(\sin(\pi)+1)=10>0$, so $f(x)$ is increasing on $(\frac{\pi}{2},\frac{3\pi}{2})$.
• For $x\in(\frac{3\pi}{2},2\pi]$, let's take $x=\frac{7\pi}{4}$. Then $f^\prime(\frac{7\pi}{4})=-10\cos(\frac{7\pi}{4})(\sin(\frac{7\pi}{4}) + 1)<0$, so $f(x)$ is decreasing on $(\frac{3\pi}{2},2\pi]$.

Step4: Find local minimum and maximum values

Evaluate $f(x)$ at the critical points $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$.

• $f(\frac{\pi}{2})=5\cos^{2}(\frac{\pi}{2})-10\sin(\frac{\pi}{2})=- 10$.
• $f(\frac{3\pi}{2})=5\cos^{2}(\frac{3\pi}{2})-10\sin(\frac{3\pi}{2})=10$. So the local minimum value is $-10$ and the local maximum value is $10$.

Step5: Find the second - derivative

Differentiate $f^\prime(x)=-10\cos(x)(\sin(x)+1)$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=-10\cos(x)$ and $v=\sin(x)+1$. $u^\prime = 10\sin(x)$ and $v^\prime=\cos(x)$. Then $f^{\prime\prime}(x)=10\sin(x)(\sin(x)+1)-10\cos^{2}(x)=10\sin^{2}(x)+10\sin(x)-10(1 - \sin^{2}(x))=20\sin^{2}(x)+10\sin(x)-10$. Set $f^{\prime\prime}(x)=0$, then $2\sin^{2}(x)+\sin(x)-1 = 0$. Let $t=\sin(x)$, so $2t^{2}+t - 1=(2t - 1)(t + 1)=0$. Then $t=\sin(x)=\frac{1}{2}$ or $\sin(x)=-1$. When $\sin(x)=\frac{1}{2}$, $x=\frac{\pi}{6},\frac{5\pi}{6}$; when $\sin(x)=-1$, $x=\frac{3\pi}{2}$. Evaluate $y = f(x)$ at these points:

• $f(\frac{\pi}{6})=5\cos^{2}(\frac{\pi}{6})-10\sin(\frac{\pi}{6})=5\times\frac{3}{4}-5=-\frac{5}{4}$.
• $f(\frac{5\pi}{6})=5\cos^{2}(\frac{5\pi}{6})-10\sin(\frac{5\pi}{6})=5\times\frac{3}{4}-5=-\frac{5}{4}$.
• $f(\frac{3\pi}{2}) = 10$. The inflection points are $(\frac{\pi}{6},-\frac{5}{4})$, $(\frac{5\pi}{6},-\frac{5}{4})$.

Step6: Determine concavity intervals

Test the intervals $[0,\frac{\pi}{6})$, $(\frac{\pi}{6},\frac{5\pi}{6})$, $(\frac{5\pi}{6},2\pi]$.

• For $x\in[0,\frac{\pi}{6})$, let $x=\frac{\pi}{12}$. Then $f^{\prime\prime}(\frac{\pi}{12})=20\sin^{2}(\frac{\pi}{12})+10\sin(\frac{\pi}{12})-10<0$, so $f(x)$ is concave down on $[0,\frac{\pi}{6})$.
• For $x\in(\frac{\pi}{6},\frac{5\pi}{6})$, let $x=\frac{\pi}{2}$. Then $f^{\prime\prime}(\frac{\pi}{2})=20\sin^{2}(\frac{\pi}{2})+10\sin(\frac{\pi}{2})-10 = 20>0$, so $f(x)$ is concave up on $(\frac{\pi}{6},\frac{5\pi}{6})$.
• For $x\in(\frac{5\pi}{6},2\pi]$, let $x=\frac{11\pi}{12}$. Then $f^{\prime\prime}(\frac{11\pi}{12})=20\sin^{2}(\frac{11\pi}{12})+10\sin(\frac{11\pi}{12})-10<0$, so $f(x)$ is concave down on $(\frac{5\pi}{6},2\pi]$.

Answer:

(a) Increasing interval: $(\frac{\pi}{2},\frac{3\pi}{2})$ Decreasing intervals: $[0,\frac{\pi}{2})\cup(\frac{3\pi}{2},2\pi]$ (b) Local minimum value: $-10$ Local maximum value: $10$ (c) Inflection points: $(\frac{\pi}{6},-\frac{5}{4})$, $(\frac{5\pi}{6},-\frac{5}{4})$ Concave up interval: $(\frac{\pi}{6},\frac{5\pi}{6})$ Concave down intervals: $[0,\frac{\pi}{6})\cup(\frac{5\pi}{6},2\pi]$