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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = \frac{12 x^{3}}{343} - \frac{54 x^{2}}{343} - \frac{360 x}{343} - \frac{65}{343}\) on \(\displaystyle [-9,6]\)

Taking the derivative gives \(\displaystyle f'(x) = \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} = 0\). The critical numbers are \(\displaystyle x = -2\) and \(\displaystyle x = 5\). The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are \(\displaystyle {-2, 5, 6, -9}\) and evaluating gives \(\displaystyle \left( -2, \ 1\right), \left( 5, \ -5\right), \left( 6, \ - \frac{1577}{343}\right), \left( -9, \ -29\right)\). The max is \(\displaystyle \left( -2, \ 1\right)\) and the min is \(\displaystyle \left( -9, \ -29\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = \frac{12 x^{3}}{343} - \frac{54 x^{2}}{343} - \frac{360 x}{343} - \frac{65}{343}$ on $[-9,6]$
    \soln{9cm}{Taking the derivative gives $f'(x) = \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343}$.  Setting it equal to zero and solving gives the critical numbers. $\frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} = 0$. The critical numbers are $x = -2$ and $x = 5$. The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are ${-2, 5, 6, -9}$ and evaluating gives $\left( -2, \  1\right), \left( 5, \  -5\right), \left( 6, \  - \frac{1577}{343}\right), \left( -9, \  -29\right)$. The max is $\left( -2, \  1\right)$ and the min is $\left( -9, \  -29\right)$.}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the absolute maximum of  <img class="equation_image" title=" \displaystyle f(x) = \frac{12 x^{3}}{343} - \frac{54 x^{2}}{343} - \frac{360 x}{343} - \frac{65}{343} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7B12%20x%5E%7B3%7D%7D%7B343%7D%20-%20%5Cfrac%7B54%20x%5E%7B2%7D%7D%7B343%7D%20-%20%5Cfrac%7B360%20x%7D%7B343%7D%20-%20%5Cfrac%7B65%7D%7B343%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{12 x^{3}}{343} - \frac{54 x^{2}}{343} - \frac{360 x}{343} - \frac{65}{343} " data-equation-content=" \displaystyle f(x) = \frac{12 x^{3}}{343} - \frac{54 x^{2}}{343} - \frac{360 x}{343} - \frac{65}{343} " />  on  <img class="equation_image" title=" \displaystyle [-9,6] " src="/equation_images/%20%5Cdisplaystyle%20%5B-9%2C6%5D%20" alt="LaTeX:  \displaystyle [-9,6] " data-equation-content=" \displaystyle [-9,6] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B36%20x%5E%7B2%7D%7D%7B343%7D%20-%20%5Cfrac%7B108%20x%7D%7B343%7D%20-%20%5Cfrac%7B360%7D%7B343%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} " data-equation-content=" \displaystyle f'(x) = \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} = 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B36%20x%5E%7B2%7D%7D%7B343%7D%20-%20%5Cfrac%7B108%20x%7D%7B343%7D%20-%20%5Cfrac%7B360%7D%7B343%7D%20%3D%200%20" alt="LaTeX:  \displaystyle \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} = 0 " data-equation-content=" \displaystyle \frac{36 x^{2}}{343} - \frac{108 x}{343} - \frac{360}{343} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " />  and  <img class="equation_image" title=" \displaystyle x = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX:  \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " /> . The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are  <img class="equation_image" title=" \displaystyle {-2, 5, 6, -9} " src="/equation_images/%20%5Cdisplaystyle%20%7B-2%2C%205%2C%206%2C%20-9%7D%20" alt="LaTeX:  \displaystyle {-2, 5, 6, -9} " data-equation-content=" \displaystyle {-2, 5, 6, -9} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( -2, \  1\right), \left( 5, \  -5\right), \left( 6, \  - \frac{1577}{343}\right), \left( -9, \  -29\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-2%2C%20%5C%20%201%5Cright%29%2C%20%5Cleft%28%205%2C%20%5C%20%20-5%5Cright%29%2C%20%5Cleft%28%206%2C%20%5C%20%20-%20%5Cfrac%7B1577%7D%7B343%7D%5Cright%29%2C%20%5Cleft%28%20-9%2C%20%5C%20%20-29%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -2, \  1\right), \left( 5, \  -5\right), \left( 6, \  - \frac{1577}{343}\right), \left( -9, \  -29\right) " data-equation-content=" \displaystyle \left( -2, \  1\right), \left( 5, \  -5\right), \left( 6, \  - \frac{1577}{343}\right), \left( -9, \  -29\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( -2, \  1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-2%2C%20%5C%20%201%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -2, \  1\right) " data-equation-content=" \displaystyle \left( -2, \  1\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( -9, \  -29\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-9%2C%20%5C%20%20-29%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -9, \  -29\right) " data-equation-content=" \displaystyle \left( -9, \  -29\right) " /> .</p> </p>