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

Please login to create an exam or a quiz.

Calculus
Applications of Derivatives
New Random

Find the local maximum and minimum of \(\displaystyle f(x) = - 3 x^{3} + x + 2\).

To find the critical numbers solve \(\displaystyle f'(x) = 0\). The derivative is \(\displaystyle f'(x) = 1 - 9 x^{2}\). Solving \(\displaystyle 1 - 9 x^{2} = 0\) gives \(\displaystyle x = \left[ - \frac{1}{3}, \ \frac{1}{3}\right]\). Using the 2nd derivative test gives:
\(\displaystyle f''\left( - \frac{1}{3} \right) = 6 \) which is greater than zero, so the function is concave up and \(\displaystyle f\left(- \frac{1}{3}\right) = \frac{16}{9}\) is a local minimum.
\(\displaystyle f''\left( \frac{1}{3} \right) = -6 \) which is less than zero, so the function is concave down and \(\displaystyle f\left(\frac{1}{3}\right) = \frac{20}{9}\) is a local maximum.

Download \(\LaTeX\) 

\begin{question}Find the local maximum and minimum of $f(x) = - 3 x^{3} + x + 2$. 
    \soln{9cm}{To find the critical numbers solve $f'(x) = 0$. The derivative is $f'(x) = 1 - 9 x^{2}$. Solving $1 - 9 x^{2} = 0$ gives $x = \left[ - \frac{1}{3}, \  \frac{1}{3}\right]$. Using the 2nd derivative test gives:\newline $f''\left( - \frac{1}{3} \right) = 6 $ which is greater than zero, so the function is concave up and $f\left(- \frac{1}{3}\right) = \frac{16}{9}$ is a local minimum.\newline $f''\left( \frac{1}{3} \right) = -6 $ which is less than zero, so the function is concave down and $f\left(\frac{1}{3}\right) = \frac{20}{9}$ is a local maximum.\newline }

\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 local maximum and minimum of  <img class="equation_image" title=" \displaystyle f(x) = - 3 x^{3} + x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%203%20x%5E%7B3%7D%20%2B%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f(x) = - 3 x^{3} + x + 2 " data-equation-content=" \displaystyle f(x) = - 3 x^{3} + x + 2 " /> . </p> </p>
HTML for Canvas
<p> <p>To find the critical numbers solve  <img class="equation_image" title=" \displaystyle f'(x) = 0 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%200%20" alt="LaTeX:  \displaystyle f'(x) = 0 " data-equation-content=" \displaystyle f'(x) = 0 " /> . The derivative is  <img class="equation_image" title=" \displaystyle f'(x) = 1 - 9 x^{2} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%201%20-%209%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle f'(x) = 1 - 9 x^{2} " data-equation-content=" \displaystyle f'(x) = 1 - 9 x^{2} " /> . Solving  <img class="equation_image" title=" \displaystyle 1 - 9 x^{2} = 0 " src="/equation_images/%20%5Cdisplaystyle%201%20-%209%20x%5E%7B2%7D%20%3D%200%20" alt="LaTeX:  \displaystyle 1 - 9 x^{2} = 0 " data-equation-content=" \displaystyle 1 - 9 x^{2} = 0 " />  gives  <img class="equation_image" title=" \displaystyle x = \left[ - \frac{1}{3}, \  \frac{1}{3}\right] " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cleft%5B%20-%20%5Cfrac%7B1%7D%7B3%7D%2C%20%5C%20%20%5Cfrac%7B1%7D%7B3%7D%5Cright%5D%20" alt="LaTeX:  \displaystyle x = \left[ - \frac{1}{3}, \  \frac{1}{3}\right] " data-equation-content=" \displaystyle x = \left[ - \frac{1}{3}, \  \frac{1}{3}\right] " /> . Using the 2nd derivative test gives:<br>  <img class="equation_image" title=" \displaystyle f''\left( - \frac{1}{3} \right) = 6  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20-%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%3D%206%20%20" alt="LaTeX:  \displaystyle f''\left( - \frac{1}{3} \right) = 6  " data-equation-content=" \displaystyle f''\left( - \frac{1}{3} \right) = 6  " />  which is greater than zero, so the function is concave up and  <img class="equation_image" title=" \displaystyle f\left(- \frac{1}{3}\right) = \frac{16}{9} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28-%20%5Cfrac%7B1%7D%7B3%7D%5Cright%29%20%3D%20%5Cfrac%7B16%7D%7B9%7D%20" alt="LaTeX:  \displaystyle f\left(- \frac{1}{3}\right) = \frac{16}{9} " data-equation-content=" \displaystyle f\left(- \frac{1}{3}\right) = \frac{16}{9} " />  is a local minimum.<br>  <img class="equation_image" title=" \displaystyle f''\left( \frac{1}{3} \right) = -6  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%3D%20-6%20%20" alt="LaTeX:  \displaystyle f''\left( \frac{1}{3} \right) = -6  " data-equation-content=" \displaystyle f''\left( \frac{1}{3} \right) = -6  " />  which is less than zero, so the function is concave down and  <img class="equation_image" title=" \displaystyle f\left(\frac{1}{3}\right) = \frac{20}{9} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%5Cright%29%20%3D%20%5Cfrac%7B20%7D%7B9%7D%20" alt="LaTeX:  \displaystyle f\left(\frac{1}{3}\right) = \frac{20}{9} " data-equation-content=" \displaystyle f\left(\frac{1}{3}\right) = \frac{20}{9} " />  is a local maximum.<br> </p> </p>