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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = - \frac{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343}\) on \(\displaystyle [-7,4]\)

Taking the derivative gives \(\displaystyle f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0\). The critical numbers are \(\displaystyle x = -5\) and \(\displaystyle x = 2\). The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are \(\displaystyle {-7, 2, -5, 4}\) and evaluating gives \(\displaystyle \left( -7, \ - \frac{1058}{343}\right), \left( 2, \ 4\right), \left( -5, \ -6\right), \left( 4, \ \frac{372}{343}\right)\). The max is \(\displaystyle \left( 2, \ 4\right)\) and the min is \(\displaystyle \left( -5, \ -6\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = - \frac{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343}$ on $[-7,4]$
    \soln{9cm}{Taking the derivative gives $f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343}$.  Setting it equal to zero and solving gives the critical numbers. $- \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0$. The critical numbers are $x = -5$ and $x = 2$. The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are ${-7, 2, -5, 4}$ and evaluating gives $\left( -7, \  - \frac{1058}{343}\right), \left( 2, \  4\right), \left( -5, \  -6\right), \left( 4, \  \frac{372}{343}\right)$. The max is $\left( 2, \  4\right)$ and the min is $\left( -5, \  -6\right)$.}

\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{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20%5Cfrac%7B20%20x%5E%7B3%7D%7D%7B343%7D%20-%20%5Cfrac%7B90%20x%5E%7B2%7D%7D%7B343%7D%20%2B%20%5Cfrac%7B600%20x%7D%7B343%7D%20%2B%20%5Cfrac%7B692%7D%7B343%7D%20" alt="LaTeX:  \displaystyle f(x) = - \frac{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343} " data-equation-content=" \displaystyle f(x) = - \frac{20 x^{3}}{343} - \frac{90 x^{2}}{343} + \frac{600 x}{343} + \frac{692}{343} " />  on  <img class="equation_image" title=" \displaystyle [-7,4] " src="/equation_images/%20%5Cdisplaystyle%20%5B-7%2C4%5D%20" alt="LaTeX:  \displaystyle [-7,4] " data-equation-content=" \displaystyle [-7,4] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B60%20x%5E%7B2%7D%7D%7B343%7D%20-%20%5Cfrac%7B180%20x%7D%7B343%7D%20%2B%20%5Cfrac%7B600%7D%7B343%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} " data-equation-content=" \displaystyle f'(x) = - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B60%20x%5E%7B2%7D%7D%7B343%7D%20-%20%5Cfrac%7B180%20x%7D%7B343%7D%20%2B%20%5Cfrac%7B600%7D%7B343%7D%20%3D%200%20" alt="LaTeX:  \displaystyle - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0 " data-equation-content=" \displaystyle - \frac{60 x^{2}}{343} - \frac{180 x}{343} + \frac{600}{343} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-5%20" alt="LaTeX:  \displaystyle x = -5 " data-equation-content=" \displaystyle x = -5 " />  and  <img class="equation_image" title=" \displaystyle x = 2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%202%20" alt="LaTeX:  \displaystyle x = 2 " data-equation-content=" \displaystyle x = 2 " /> . 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 {-7, 2, -5, 4} " src="/equation_images/%20%5Cdisplaystyle%20%7B-7%2C%202%2C%20-5%2C%204%7D%20" alt="LaTeX:  \displaystyle {-7, 2, -5, 4} " data-equation-content=" \displaystyle {-7, 2, -5, 4} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( -7, \  - \frac{1058}{343}\right), \left( 2, \  4\right), \left( -5, \  -6\right), \left( 4, \  \frac{372}{343}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-7%2C%20%5C%20%20-%20%5Cfrac%7B1058%7D%7B343%7D%5Cright%29%2C%20%5Cleft%28%202%2C%20%5C%20%204%5Cright%29%2C%20%5Cleft%28%20-5%2C%20%5C%20%20-6%5Cright%29%2C%20%5Cleft%28%204%2C%20%5C%20%20%5Cfrac%7B372%7D%7B343%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -7, \  - \frac{1058}{343}\right), \left( 2, \  4\right), \left( -5, \  -6\right), \left( 4, \  \frac{372}{343}\right) " data-equation-content=" \displaystyle \left( -7, \  - \frac{1058}{343}\right), \left( 2, \  4\right), \left( -5, \  -6\right), \left( 4, \  \frac{372}{343}\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 2, \  4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%204%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  4\right) " data-equation-content=" \displaystyle \left( 2, \  4\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( -5, \  -6\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-5%2C%20%5C%20%20-6%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -5, \  -6\right) " data-equation-content=" \displaystyle \left( -5, \  -6\right) " /> .</p> </p>