quiz 6 - requires respondus lockdown browser + webcam\nstarted: sep 27 at 12:25pm\nquiz instructions\naccess…

quiz 6 - requires respondus lockdown browser + webcam\nstarted: sep 27 at 12:25pm\nquiz instructions\naccess code: start\ntimed: 30 minutes\nnumber of attempts: 1\nlockdown browser and respondus monitor required.\nthis quiz covers material from this week (sections 2.7, 2.8, 2.9).\nquestion 5\nfind the linearization, $l(x)$, of the function at $a$.\n$f(x)=sin(x)$ where $a = \frac{pi}{6}$.\n$\frac{sqrt{3}x}{2}-\frac{pi}{4sqrt{3}}+\frac{1}{2}$\nthe correct answer is not listed.\n$\frac{2x}{3}+\frac{pi}{4sqrt{3}}+\frac{1}{3}$\n$\frac{sqrt{3}x}{2}+\frac{pi}{4sqrt{2}}+\frac{1}{2}$

quiz 6 - requires respondus lockdown browser + webcam\nstarted: sep 27 at 12:25pm\nquiz instructions\naccess code: start\ntimed: 30 minutes\nnumber of attempts: 1\nlockdown browser and respondus monitor required.\nthis quiz covers material from this week (sections 2.7, 2.8, 2.9).\nquestion 5\nfind the linearization, $l(x)$, of the function at $a$.\n$f(x)=sin(x)$ where $a = \frac{pi}{6}$.\n$\frac{sqrt{3}x}{2}-\frac{pi}{4sqrt{3}}+\frac{1}{2}$\nthe correct answer is not listed.\n$\frac{2x}{3}+\frac{pi}{4sqrt{3}}+\frac{1}{3}$\n$\frac{sqrt{3}x}{2}+\frac{pi}{4sqrt{2}}+\frac{1}{2}$

Answer

Explanation:

Step1: Recall linearization formula

The linearization of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$.

Step2: Find $f(a)$

Given $f(x)=\sin(x)$ and $a=\frac{\pi}{6}$, then $f(a)=\sin(\frac{\pi}{6})=\frac{1}{2}$.

Step3: Find $f^{\prime}(x)$ and $f^{\prime}(a)$

The derivative of $f(x)=\sin(x)$ is $f^{\prime}(x)=\cos(x)$. So $f^{\prime}(a)=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$.

Step4: Substitute into linearization formula

$L(x)=f(\frac{\pi}{6})+f^{\prime}(\frac{\pi}{6})(x - \frac{\pi}{6})=\frac{1}{2}+\frac{\sqrt{3}}{2}(x-\frac{\pi}{6})=\frac{\sqrt{3}x}{2}-\frac{\pi\sqrt{3}}{12}+\frac{1}{2}=\frac{\sqrt{3}x}{2}-\frac{\pi}{4\sqrt{3}}+\frac{1}{2}$.

Answer:

$\frac{\sqrt{3}x}{2}-\frac{\pi}{4\sqrt{3}}+\frac{1}{2}$