Question

quiz 6 - requires respondus lockdown browser + webcam
started: sep 27 at 12:25pm
quiz instructions
access code: start
timed: 30 minutes
number of attempts: 1
lockdown browser and respondus monitor required.
this quiz covers material from this week (sections 2.7, 2.8, 2.9).
question 5
find the linearization, $l(x)$, of the function at $a$.
$f(x)=sin(x)$ where $a = \frac{pi}{6}$.
$\frac{sqrt{3}x}{2}-\frac{pi}{4sqrt{3}}+\frac{1}{2}$
the correct answer is not listed.
$\frac{2x}{3}+\frac{pi}{4sqrt{3}}+\frac{1}{3}$
$\frac{sqrt{3}x}{2}+\frac{pi}{4sqrt{2}}+\frac{1}{2}$

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:

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