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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = \frac{22 x^{3}}{729} - \frac{55 x^{2}}{243} - \frac{308 x}{243} + \frac{4820}{729}\) on \(\displaystyle [-5,7]\)

Taking the derivative gives \(\displaystyle f'(x) = \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} = 0\). The critical numbers are \(\displaystyle x = -2\) and \(\displaystyle x = 7\). The absolute maximum is either at a critical number or at the end point of the interval. The critical number \(\displaystyle x = 7\) is not in the interval \(\displaystyle (-5,7)\) and does not need to be checked. The inputs to be checked are \(\displaystyle {-5, -2, 7}\) and evaluating gives \(\displaystyle \left( -5, \ \frac{95}{27}\right), \left( -2, \ 8\right), \left( 7, \ -3\right)\). The max is \(\displaystyle \left( -2, \ 8\right)\) and the min is \(\displaystyle \left( 7, \ -3\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = \frac{22 x^{3}}{729} - \frac{55 x^{2}}{243} - \frac{308 x}{243} + \frac{4820}{729}$ on $[-5,7]$
    \soln{9cm}{Taking the derivative gives $f'(x) = \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243}$.  Setting it equal to zero and solving gives the critical numbers. $\frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} = 0$. The critical numbers are $x = -2$ and $x = 7$. The absolute maximum is either at a critical number or at the end point of the interval. The critical number $x = 7$ is not in the interval $(-5,7)$ and does not need to be checked. The inputs to be checked are ${-5, -2, 7}$ and evaluating gives $\left( -5, \  \frac{95}{27}\right), \left( -2, \  8\right), \left( 7, \  -3\right)$. The max is $\left( -2, \  8\right)$ and the min is $\left( 7, \  -3\right)$.}

\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{22 x^{3}}{729} - \frac{55 x^{2}}{243} - \frac{308 x}{243} + \frac{4820}{729} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7B22%20x%5E%7B3%7D%7D%7B729%7D%20-%20%5Cfrac%7B55%20x%5E%7B2%7D%7D%7B243%7D%20-%20%5Cfrac%7B308%20x%7D%7B243%7D%20%2B%20%5Cfrac%7B4820%7D%7B729%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{22 x^{3}}{729} - \frac{55 x^{2}}{243} - \frac{308 x}{243} + \frac{4820}{729} " data-equation-content=" \displaystyle f(x) = \frac{22 x^{3}}{729} - \frac{55 x^{2}}{243} - \frac{308 x}{243} + \frac{4820}{729} " />  on  <img class="equation_image" title=" \displaystyle [-5,7] " src="/equation_images/%20%5Cdisplaystyle%20%5B-5%2C7%5D%20" alt="LaTeX:  \displaystyle [-5,7] " data-equation-content=" \displaystyle [-5,7] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B22%20x%5E%7B2%7D%7D%7B243%7D%20-%20%5Cfrac%7B110%20x%7D%7B243%7D%20-%20%5Cfrac%7B308%7D%7B243%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} " data-equation-content=" \displaystyle f'(x) = \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} = 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B22%20x%5E%7B2%7D%7D%7B243%7D%20-%20%5Cfrac%7B110%20x%7D%7B243%7D%20-%20%5Cfrac%7B308%7D%7B243%7D%20%3D%200%20" alt="LaTeX:  \displaystyle \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} = 0 " data-equation-content=" \displaystyle \frac{22 x^{2}}{243} - \frac{110 x}{243} - \frac{308}{243} = 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 = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX:  \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " /> . 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 = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX:  \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " />  is not in the interval  <img class="equation_image" title=" \displaystyle (-5,7) " src="/equation_images/%20%5Cdisplaystyle%20%28-5%2C7%29%20" alt="LaTeX:  \displaystyle (-5,7) " data-equation-content=" \displaystyle (-5,7) " />  and does not need to be checked. The inputs to be checked are  <img class="equation_image" title=" \displaystyle {-5, -2, 7} " src="/equation_images/%20%5Cdisplaystyle%20%7B-5%2C%20-2%2C%207%7D%20" alt="LaTeX:  \displaystyle {-5, -2, 7} " data-equation-content=" \displaystyle {-5, -2, 7} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( -5, \  \frac{95}{27}\right), \left( -2, \  8\right), \left( 7, \  -3\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-5%2C%20%5C%20%20%5Cfrac%7B95%7D%7B27%7D%5Cright%29%2C%20%5Cleft%28%20-2%2C%20%5C%20%208%5Cright%29%2C%20%5Cleft%28%207%2C%20%5C%20%20-3%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -5, \  \frac{95}{27}\right), \left( -2, \  8\right), \left( 7, \  -3\right) " data-equation-content=" \displaystyle \left( -5, \  \frac{95}{27}\right), \left( -2, \  8\right), \left( 7, \  -3\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( -2, \  8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-2%2C%20%5C%20%208%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -2, \  8\right) " data-equation-content=" \displaystyle \left( -2, \  8\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( 7, \  -3\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%207%2C%20%5C%20%20-3%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 7, \  -3\right) " data-equation-content=" \displaystyle \left( 7, \  -3\right) " /> .</p> </p>