\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the value of \(\displaystyle c\) for \(\displaystyle f(x)=5 x^{2} - 5 x + 6\) that satisfies the Mean Value Theorem on \(\displaystyle [-9, 3]\).
The slope of the secant line is \(\displaystyle m = \frac{f(3) -f(-9)}{3-(-9)}=-35\). Setting the derivative equal to the slope of the secant line gives the equation \(\displaystyle -35 = 10 x - 5\). The solution is \(\displaystyle c = -3\).
\begin{question}Find the value of $c$ for $f(x)=5 x^{2} - 5 x + 6$ that satisfies the Mean Value Theorem on $[-9, 3]$.
\soln{9cm}{The slope of the secant line is $m = \frac{f(3) -f(-9)}{3-(-9)}=-35$. Setting the derivative equal to the slope of the secant line gives the equation $-35 = 10 x - 5$. The solution is $c = -3$. }
\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)=5 x^{2} - 5 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D5%20x%5E%7B2%7D%20-%205%20x%20%2B%206%20" alt="LaTeX: \displaystyle f(x)=5 x^{2} - 5 x + 6 " data-equation-content=" \displaystyle f(x)=5 x^{2} - 5 x + 6 " /> that satisfies the Mean Value Theorem on <img class="equation_image" title=" \displaystyle [-9, 3] " src="/equation_images/%20%5Cdisplaystyle%20%5B-9%2C%203%5D%20" alt="LaTeX: \displaystyle [-9, 3] " data-equation-content=" \displaystyle [-9, 3] " /> . </p> </p><p> <p>The slope of the secant line is <img class="equation_image" title=" \displaystyle m = \frac{f(3) -f(-9)}{3-(-9)}=-35 " src="/equation_images/%20%5Cdisplaystyle%20m%20%3D%20%5Cfrac%7Bf%283%29%20-f%28-9%29%7D%7B3-%28-9%29%7D%3D-35%20" alt="LaTeX: \displaystyle m = \frac{f(3) -f(-9)}{3-(-9)}=-35 " data-equation-content=" \displaystyle m = \frac{f(3) -f(-9)}{3-(-9)}=-35 " /> . Setting the derivative equal to the slope of the secant line gives the equation <img class="equation_image" title=" \displaystyle -35 = 10 x - 5 " src="/equation_images/%20%5Cdisplaystyle%20-35%20%3D%2010%20x%20-%205%20" alt="LaTeX: \displaystyle -35 = 10 x - 5 " data-equation-content=" \displaystyle -35 = 10 x - 5 " /> . The solution is <img class="equation_image" title=" \displaystyle c = -3 " src="/equation_images/%20%5Cdisplaystyle%20c%20%3D%20-3%20" alt="LaTeX: \displaystyle c = -3 " data-equation-content=" \displaystyle c = -3 " /> . </p> </p>