Question

instructions

• work this problem out on 1 sheet of paper, one - side only. label the problem, \linearization.\
• you will make a pdf of all of your work - outs after you complete the test;
• make this problem page 3;
• after you submit this test, then submit your pdf in the canvas portal titled, \exam 2 work - out questions\
• an instructor will grade your work - out questions.

work - out problem 3, 10 points
find the linearization $l(x)$ of the function $f(x)=sin(x)$, at $x = \frac{pi}{3}$. the answer is graded on the work you present. the linearization must consist entirely of exact values (reduced fractions of integers, values expressed as square roots where appropriate). no credit for decimal expressions and/or approximations.
click to proceed when you have completed the work on your own paper.

Explanation:

Step1: Recall linearization formula

The linearization $L(x)$ 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 $a=\frac{\pi}{3}$, then $f(a)=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$.

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

The derivative of $f(x)=\sin(x)$ is $f^{\prime}(x)=\cos(x)$.

Step4: Find $f^{\prime}(a)$

Substitute $a = \frac{\pi}{3}$ into $f^{\prime}(x)$, so $f^{\prime}(\frac{\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}$.

Step5: Find $L(x)$

Substitute $f(a)=\frac{\sqrt{3}}{2}$, $f^{\prime}(a)=\frac{1}{2}$ and $a=\frac{\pi}{3}$ into the linearization formula:
\[

$$\begin{align*} L(x)&=\frac{\sqrt{3}}{2}+\frac{1}{2}(x-\frac{\pi}{3})\\ &=\frac{1}{2}x+\frac{\sqrt{3}}{2}-\frac{\pi}{6} \end{align*}$$

\]

Answer:

$L(x)=\frac{1}{2}x+\frac{\sqrt{3}}{2}-\frac{\pi}{6}$