Generate Math Exams and Quizzes in LaTeX and pdf formats

Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}} \end{equation*} Using \(\displaystyle x_0 = 3\) and \(\displaystyle n = 0,1,2,3,\) and \(\displaystyle 4\) gives: \begin{equation*}x_{1} = (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109\end{equation*} \begin{equation*}x_{2} = (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173\end{equation*} \begin{equation*}x_{3} = (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721\end{equation*} \begin{equation*}x_{4} = (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785\end{equation*} \begin{equation*}x_{5} = (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785\end{equation*}

Download \(\LaTeX\)

\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{68 x^{3}}{125} - 9$ using $x_0=3$. 
    \soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} =  x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}}  \end{equation*}
Using $x_0 = 3$ and $n = 0,1,2,3,$ and $4$ gives:
\begin{equation*}x_{1} =  (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109\end{equation*}
\begin{equation*}x_{2} =  (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173\end{equation*}
\begin{equation*}x_{3} =  (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721\end{equation*}
\begin{equation*}x_{4} =  (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785\end{equation*}
\begin{equation*}x_{5} =  (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785\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>Use Newton's method to find the first 5 approximations of the solution to the equation  <img class="equation_image" title=" \displaystyle e^{- x}= \frac{68 x^{3}}{125} - 9 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B68%20x%5E%7B3%7D%7D%7B125%7D%20-%209%20" alt="LaTeX:  \displaystyle e^{- x}= \frac{68 x^{3}}{125} - 9 " data-equation-content=" \displaystyle e^{- x}= \frac{68 x^{3}}{125} - 9 " />  using  <img class="equation_image" title=" \displaystyle x_0=3 " src="/equation_images/%20%5Cdisplaystyle%20x_0%3D3%20" alt="LaTeX:  \displaystyle x_0=3 " data-equation-content=" \displaystyle x_0=3 " /> . </p> </p>

HTML for Canvas

<p> <p>Using the formula for Newton's method gives
 <img class="equation_image" title=" x_{n+1} =  x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}}   " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20x_%7Bn%7D%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B204%20x_%7Bn%7D%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="LaTeX:  x_{n+1} =  x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}}   " data-equation-content=" x_{n+1} =  x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}}   " /> 
Using  <img class="equation_image" title=" \displaystyle x_0 = 3 " src="/equation_images/%20%5Cdisplaystyle%20x_0%20%3D%203%20" alt="LaTeX:  \displaystyle x_0 = 3 " data-equation-content=" \displaystyle x_0 = 3 " />  and  <img class="equation_image" title=" \displaystyle n = 0,1,2,3, " src="/equation_images/%20%5Cdisplaystyle%20n%20%3D%200%2C1%2C2%2C3%2C%20" alt="LaTeX:  \displaystyle n = 0,1,2,3, " data-equation-content=" \displaystyle n = 0,1,2,3, " />  and  <img class="equation_image" title=" \displaystyle 4 " src="/equation_images/%20%5Cdisplaystyle%204%20" alt="LaTeX:  \displaystyle 4 " data-equation-content=" \displaystyle 4 " />  gives:
 <img class="equation_image" title=" x_{1} =  (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%283.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20%283.0000000000%29%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20%283.0000000000%29%7D%7D%7B-%20%5Cfrac%7B204%20%283.0000000000%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%283.0000000000%29%7D%7D%20%3D%202.6174315109%20" alt="LaTeX:  x_{1} =  (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109 " data-equation-content=" x_{1} =  (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109 " /> 
 <img class="equation_image" title=" x_{2} =  (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.6174315109%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20%282.6174315109%29%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20%282.6174315109%29%7D%7D%7B-%20%5Cfrac%7B204%20%282.6174315109%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%282.6174315109%29%7D%7D%20%3D%202.5568333173%20" alt="LaTeX:  x_{2} =  (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173 " data-equation-content=" x_{2} =  (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173 " /> 
 <img class="equation_image" title=" x_{3} =  (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.5568333173%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20%282.5568333173%29%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20%282.5568333173%29%7D%7D%7B-%20%5Cfrac%7B204%20%282.5568333173%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%282.5568333173%29%7D%7D%20%3D%202.5553976721%20" alt="LaTeX:  x_{3} =  (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721 " data-equation-content=" x_{3} =  (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721 " /> 
 <img class="equation_image" title=" x_{4} =  (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%282.5553976721%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20%282.5553976721%29%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20%282.5553976721%29%7D%7D%7B-%20%5Cfrac%7B204%20%282.5553976721%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%282.5553976721%29%7D%7D%20%3D%202.5553968785%20" alt="LaTeX:  x_{4} =  (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785 " data-equation-content=" x_{4} =  (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785 " /> 
 <img class="equation_image" title=" x_{5} =  (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%282.5553968785%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B68%20%282.5553968785%29%5E%7B3%7D%7D%7B125%7D%20%2B%209%20%2B%20e%5E%7B-%20%282.5553968785%29%7D%7D%7B-%20%5Cfrac%7B204%20%282.5553968785%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%282.5553968785%29%7D%7D%20%3D%202.5553968785%20" alt="LaTeX:  x_{5} =  (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785 " data-equation-content=" x_{5} =  (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785 " /> 
</p> </p>