Math 181 Miniproject 3: Texting Lesson.md --- My lesson Topic === <style> body { background-color: #eeeeee; } h1 { color: maroon; margin-left: 40px; } .gray { margin-left: 50px ; margin-right: 29%; font-weight: 500; color: #000000; background-color: #cccccc; border-color: #aaaaaa; } .blue { display: inline-block; margin-left: 29% ; margin-right: 0%; width: -webkit-calc(70% - 50px); width: -moz-calc(70% - 50px); width: calc(70% - 50px); font-weight: 500; color: #fff; border-color: #336699; background-color: #337799; } .left { content:url("https://i.imgur.com/rUsxo7j.png"); width:50px; border-radius: 50%; float:left; } .right{ content:url("https://i.imgur.com/5ALcyl3.png"); width:50px; border-radius: 50%; display: inline-block; vertical-align:top; } </style> <div id="container" style=" padding: 6px; color: #fff; border-color: #336699; background-color: #337799; display: flex; justify-content: space-between; margin-bottom:3px;"> <div> <i class="fa fa-envelope fa-2x"></i> </div> <div> <i class="fa fa-camera fa-2x"></i> </div> <div> <i class="fa fa-comments fa-2x"></i> </div> <div> <i class="fa fa-address-card fa-2x" aria-hidden="true"></i> </div> <div> <i class="fa fa-phone fa-2x" aria-hidden="true"></i> </div> <div> <i class="fa fa-list-ul fa-2x" aria-hidden="true"></i> </div> <div> <i class="fa fa-user-plus fa-2x" aria-hidden="true"></i> </div> </div> <div><img class="left"/><div class="alert gray"> Hey can you explain 1.8 on how to use the limit derivative formula to find the local linearization and how to use L(x) to estimate the value? My question says for the function $f(x)=x^6$ you are given $f'(x)=6x^5$ find the local linearization L(x) of f(x) at x=2. Use L(x) to estimate the value of 2.02. </div></div> <div><div class="alert blue"> Yes! You are going to determine the slope of the tangent line $y=f(x)$ at $(a,f(a))$ by computing $f'(a)$. The equation you're going to use is the tangent line given in poin-slope form. Point-slope form is $y-y_1=m(x-x_1)$ but for the tangent line, it looks a little bit different $y-f(a)=f'(a)(x-a)$ add $-f(a)$ to both sides giving you a new equation of $y=f(a)+f'(a)(x-a)$. You then will replace y with $L(x)$. Your equation will look like $L(x)=f(a)+f'(a)(x-a)$. </div><img class="right"/></div> </div></div> <div><div class="alert blue"> Your equation is for the function $f(x)=x^6$ you are given $f'(x)=6x^5$ You have to find the local linearization $L(x)$ of $f(x)$ at $x=2$ Use the $L(x)=f(a)+f'(a)(x-a)$ equation 1) Plug $x=2$ in for a. Your equation will look like $L(x)=f(2)+f'(2)(x-2)$. 2) Plug 2 in for x for $f(x) and f'(x)$. Your quation will look like $L(x)=f(2)+f'(2)(x-2)$ now plug 2 into $f(x) and f'(x)$. The equation will look like $=(2)^6 +6(2)^5(x-2)$ 3) Simplify $(2)^6$ and $6(2)^5$ You get 64 and 192 4) Plug them back into the equation like this $=64+192(x-2)$ 5) Now solve $192(x-2)$ which will give you $192x-384$ 6) Your final answer will be $64+192x-384$ </div><img class="right"/></div> </div></div> <div><div class="alert blue"> The second part of your question is use $L(x)$ to estimate the value of 2.02 You are going to use the same $L(x)=f(a)+f'(a)(x-1)$ equation but you don't have to solve for $f(a)+f'(x)(x-1)$ because you already did it in part one. 1) Plug 2.02 in for x $L(2.02)=64+192(2.02)-384$ 2) Now simplify and your answer will be 67.84 </div><img class="right"/></div> <div><img class="left"/><div class="alert gray"> That really helped, thank you. What would you suggest for studying when it comes time for an exam? </div></div> <div><div class="alert blue"> I would say for extra studying for the exam, rewatch the lecture recording on 1.8. The professor gives a lot of examples on it and you can also go to the homework and click on "practice" and it will give you different types of practice problems. Once you memorize the equation all you have to do is plug in the numbers. You can also look up videos of different problems! </div><img class="right"/></div> --- To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.