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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = - \frac{4 x^{3}}{1331} - \frac{42 x^{2}}{1331} + \frac{216 x}{1331} + \frac{13078}{1331}\) on \(\displaystyle [-9,2]\)

Taking the derivative gives \(\displaystyle f'(x) = - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} = 0\). The critical numbers are \(\displaystyle x = -9\) and \(\displaystyle x = 2\). The absolute maximum is either at a critical number or at the end point of the interval. The critical number \(\displaystyle x = -9\) is not in the interval \(\displaystyle (-9,2)\) and does not need to be checked. The critical number \(\displaystyle x = 2\) is not in the interval \(\displaystyle (-9,2)\) and does not need to be checked. The inputs to be checked are \(\displaystyle {2, -9}\) and evaluating gives \(\displaystyle \left( 2, \ 10\right), \left( -9, \ 8\right)\). The max is \(\displaystyle \left( 2, \ 10\right)\) and the min is \(\displaystyle \left( -9, \ 8\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = - \frac{4 x^{3}}{1331} - \frac{42 x^{2}}{1331} + \frac{216 x}{1331} + \frac{13078}{1331}$ on $[-9,2]$
    \soln{9cm}{Taking the derivative gives $f'(x) = - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331}$.  Setting it equal to zero and solving gives the critical numbers. $- \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} = 0$. The critical numbers are $x = -9$ and $x = 2$. The absolute maximum is either at a critical number or at the end point of the interval. The critical number $x = -9$ is not in the interval $(-9,2)$ and does not need to be checked. The critical number $x = 2$ is not in the interval $(-9,2)$ and does not need to be checked. The inputs to be checked are ${2, -9}$ and evaluating gives $\left( 2, \  10\right), \left( -9, \  8\right)$. The max is $\left( 2, \  10\right)$ and the min is $\left( -9, \  8\right)$.}

\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{4 x^{3}}{1331} - \frac{42 x^{2}}{1331} + \frac{216 x}{1331} + \frac{13078}{1331} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20%5Cfrac%7B4%20x%5E%7B3%7D%7D%7B1331%7D%20-%20%5Cfrac%7B42%20x%5E%7B2%7D%7D%7B1331%7D%20%2B%20%5Cfrac%7B216%20x%7D%7B1331%7D%20%2B%20%5Cfrac%7B13078%7D%7B1331%7D%20" alt="LaTeX:  \displaystyle f(x) = - \frac{4 x^{3}}{1331} - \frac{42 x^{2}}{1331} + \frac{216 x}{1331} + \frac{13078}{1331} " data-equation-content=" \displaystyle f(x) = - \frac{4 x^{3}}{1331} - \frac{42 x^{2}}{1331} + \frac{216 x}{1331} + \frac{13078}{1331} " />  on  <img class="equation_image" title=" \displaystyle [-9,2] " src="/equation_images/%20%5Cdisplaystyle%20%5B-9%2C2%5D%20" alt="LaTeX:  \displaystyle [-9,2] " data-equation-content=" \displaystyle [-9,2] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B12%20x%5E%7B2%7D%7D%7B1331%7D%20-%20%5Cfrac%7B84%20x%7D%7B1331%7D%20%2B%20%5Cfrac%7B216%7D%7B1331%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} " data-equation-content=" \displaystyle f'(x) = - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B12%20x%5E%7B2%7D%7D%7B1331%7D%20-%20%5Cfrac%7B84%20x%7D%7B1331%7D%20%2B%20%5Cfrac%7B216%7D%7B1331%7D%20%3D%200%20" alt="LaTeX:  \displaystyle - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} = 0 " data-equation-content=" \displaystyle - \frac{12 x^{2}}{1331} - \frac{84 x}{1331} + \frac{216}{1331} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -9 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-9%20" alt="LaTeX:  \displaystyle x = -9 " data-equation-content=" \displaystyle x = -9 " />  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 critical number  <img class="equation_image" title=" \displaystyle x = -9 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-9%20" alt="LaTeX:  \displaystyle x = -9 " data-equation-content=" \displaystyle x = -9 " />  is not in the interval  <img class="equation_image" title=" \displaystyle (-9,2) " src="/equation_images/%20%5Cdisplaystyle%20%28-9%2C2%29%20" alt="LaTeX:  \displaystyle (-9,2) " data-equation-content=" \displaystyle (-9,2) " />  and does not need to be checked. The critical number  <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 " />  is not in the interval  <img class="equation_image" title=" \displaystyle (-9,2) " src="/equation_images/%20%5Cdisplaystyle%20%28-9%2C2%29%20" alt="LaTeX:  \displaystyle (-9,2) " data-equation-content=" \displaystyle (-9,2) " />  and does not need to be checked. The inputs to be checked are  <img class="equation_image" title=" \displaystyle {2, -9} " src="/equation_images/%20%5Cdisplaystyle%20%7B2%2C%20-9%7D%20" alt="LaTeX:  \displaystyle {2, -9} " data-equation-content=" \displaystyle {2, -9} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 2, \  10\right), \left( -9, \  8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%2010%5Cright%29%2C%20%5Cleft%28%20-9%2C%20%5C%20%208%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  10\right), \left( -9, \  8\right) " data-equation-content=" \displaystyle \left( 2, \  10\right), \left( -9, \  8\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 2, \  10\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%2010%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  10\right) " data-equation-content=" \displaystyle \left( 2, \  10\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( -9, \  8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-9%2C%20%5C%20%208%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -9, \  8\right) " data-equation-content=" \displaystyle \left( -9, \  8\right) " /> .</p> </p>