\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Calculus
Applications of Derivatives
New Random

Find the linear approximation of \(\displaystyle f(x) = \sqrt{x + 144}\) at \(\displaystyle a = 0\) and use it to approximate \(\displaystyle \sqrt{144.48}\)

Using the formula for the linearization \(\displaystyle L(x) = f'(a)(x-a)+f(a)\) gives: \begin{equation*}f(x) = \sqrt{x + 144} \approx L(x) = \frac{x}{24} + 12 \end{equation*} To approximate \(\displaystyle \sqrt{144.48}\) use \(\displaystyle x = 144.48-144 = 0.48\). \begin{equation*}\sqrt{144.48} = f(0.48) \approx L(0.48) = 12.020000 \end{equation*}

Download \(\LaTeX\) 

\begin{question}Find the linear approximation of $f(x) = \sqrt{x + 144}$ at $a = 0$ and use it to approximate $\sqrt{144.48}$
    \soln{9cm}{Using the formula for the linearization $L(x) = f'(a)(x-a)+f(a)$ gives:
\begin{equation*}f(x) = \sqrt{x + 144} \approx L(x) = \frac{x}{24} + 12 \end{equation*}
To approximate $\sqrt{144.48}$ use $x = 144.48-144 = 0.48$.
\begin{equation*}\sqrt{144.48} = f(0.48)  \approx L(0.48) = 12.020000 \end{equation*}
}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}
HTML for Canvas 
<p> <p>Find the linear approximation of  <img class="equation_image" title=" \displaystyle f(x) = \sqrt{x + 144} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20144%7D%20" alt="LaTeX:  \displaystyle f(x) = \sqrt{x + 144} " data-equation-content=" \displaystyle f(x) = \sqrt{x + 144} " />  at  <img class="equation_image" title=" \displaystyle a = 0 " src="/equation_images/%20%5Cdisplaystyle%20a%20%3D%200%20" alt="LaTeX:  \displaystyle a = 0 " data-equation-content=" \displaystyle a = 0 " />  and use it to approximate  <img class="equation_image" title=" \displaystyle \sqrt{144.48} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B144.48%7D%20" alt="LaTeX:  \displaystyle \sqrt{144.48} " data-equation-content=" \displaystyle \sqrt{144.48} " /> </p> </p>
HTML for Canvas
<p> <p>Using the formula for the linearization  <img class="equation_image" title=" \displaystyle L(x) = f'(a)(x-a)+f(a) " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%20%3D%20f%27%28a%29%28x-a%29%2Bf%28a%29%20" alt="LaTeX:  \displaystyle L(x) = f'(a)(x-a)+f(a) " data-equation-content=" \displaystyle L(x) = f'(a)(x-a)+f(a) " />  gives:
 <img class="equation_image" title=" f(x) = \sqrt{x + 144} \approx L(x) = \frac{x}{24} + 12  " src="/equation_images/%20f%28x%29%20%3D%20%5Csqrt%7Bx%20%2B%20144%7D%20%5Capprox%20L%28x%29%20%3D%20%5Cfrac%7Bx%7D%7B24%7D%20%2B%2012%20%20" alt="LaTeX:  f(x) = \sqrt{x + 144} \approx L(x) = \frac{x}{24} + 12  " data-equation-content=" f(x) = \sqrt{x + 144} \approx L(x) = \frac{x}{24} + 12  " /> 
To approximate  <img class="equation_image" title=" \displaystyle \sqrt{144.48} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B144.48%7D%20" alt="LaTeX:  \displaystyle \sqrt{144.48} " data-equation-content=" \displaystyle \sqrt{144.48} " />  use  <img class="equation_image" title=" \displaystyle x = 144.48-144 = 0.48 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20144.48-144%20%3D%200.48%20" alt="LaTeX:  \displaystyle x = 144.48-144 = 0.48 " data-equation-content=" \displaystyle x = 144.48-144 = 0.48 " /> .
 <img class="equation_image" title=" \sqrt{144.48} = f(0.48)  \approx L(0.48) = 12.020000  " src="/equation_images/%20%5Csqrt%7B144.48%7D%20%3D%20f%280.48%29%20%20%5Capprox%20L%280.48%29%20%3D%2012.020000%20%20" alt="LaTeX:  \sqrt{144.48} = f(0.48)  \approx L(0.48) = 12.020000  " data-equation-content=" \sqrt{144.48} = f(0.48)  \approx L(0.48) = 12.020000  " /> 
</p> </p>