\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the value of \(\displaystyle c\) for \(\displaystyle f(x)=- 8 x^{2} + 3 x - 6\) that satisfies the Mean Value Theorem on \(\displaystyle [-7, 4]\).
The slope of the secant line is \(\displaystyle m = \frac{f(4) -f(-7)}{4-(-7)}=27\). Setting the derivative equal to the slope of the secant line gives the equation \(\displaystyle 27 = 3 - 16 x\). The solution is \(\displaystyle c = - \frac{3}{2}\).
\begin{question}Find the value of $c$ for $f(x)=- 8 x^{2} + 3 x - 6$ that satisfies the Mean Value Theorem on $[-7, 4]$.
\soln{9cm}{The slope of the secant line is $m = \frac{f(4) -f(-7)}{4-(-7)}=27$. Setting the derivative equal to the slope of the secant line gives the equation $27 = 3 - 16 x$. The solution is $c = - \frac{3}{2}$. }
\end{question}
\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}<p> <p>Find the value of <img class="equation_image" title=" \displaystyle c " src="/equation_images/%20%5Cdisplaystyle%20c%20" alt="LaTeX: \displaystyle c " data-equation-content=" \displaystyle c " /> for <img class="equation_image" title=" \displaystyle f(x)=- 8 x^{2} + 3 x - 6 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D-%208%20x%5E%7B2%7D%20%2B%203%20x%20-%206%20" alt="LaTeX: \displaystyle f(x)=- 8 x^{2} + 3 x - 6 " data-equation-content=" \displaystyle f(x)=- 8 x^{2} + 3 x - 6 " /> that satisfies the Mean Value Theorem on <img class="equation_image" title=" \displaystyle [-7, 4] " src="/equation_images/%20%5Cdisplaystyle%20%5B-7%2C%204%5D%20" alt="LaTeX: \displaystyle [-7, 4] " data-equation-content=" \displaystyle [-7, 4] " /> . </p> </p><p> <p>The slope of the secant line is <img class="equation_image" title=" \displaystyle m = \frac{f(4) -f(-7)}{4-(-7)}=27 " src="/equation_images/%20%5Cdisplaystyle%20m%20%3D%20%5Cfrac%7Bf%284%29%20-f%28-7%29%7D%7B4-%28-7%29%7D%3D27%20" alt="LaTeX: \displaystyle m = \frac{f(4) -f(-7)}{4-(-7)}=27 " data-equation-content=" \displaystyle m = \frac{f(4) -f(-7)}{4-(-7)}=27 " /> . Setting the derivative equal to the slope of the secant line gives the equation <img class="equation_image" title=" \displaystyle 27 = 3 - 16 x " src="/equation_images/%20%5Cdisplaystyle%2027%20%3D%203%20-%2016%20x%20" alt="LaTeX: \displaystyle 27 = 3 - 16 x " data-equation-content=" \displaystyle 27 = 3 - 16 x " /> . The solution is <img class="equation_image" title=" \displaystyle c = - \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20c%20%3D%20-%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle c = - \frac{3}{2} " data-equation-content=" \displaystyle c = - \frac{3}{2} " /> . </p> </p>