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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = - \frac{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125}\) on \(\displaystyle [-2,10]\)

Taking the derivative gives \(\displaystyle f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0\). The critical numbers are \(\displaystyle x = -1\) and \(\displaystyle x = 4\). The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are \(\displaystyle {10, 4, -2, -1}\) and evaluating gives \(\displaystyle \left( 10, \ - \frac{3888}{125}\right), \left( 4, \ 0\right), \left( -2, \ - \frac{432}{125}\right), \left( -1, \ -4\right)\). The max is \(\displaystyle \left( 4, \ 0\right)\) and the min is \(\displaystyle \left( 10, \ - \frac{3888}{125}\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = - \frac{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125}$ on $[-2,10]$
    \soln{9cm}{Taking the derivative gives $f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125}$.  Setting it equal to zero and solving gives the critical numbers. $- \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0$. The critical numbers are $x = -1$ and $x = 4$. The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are ${10, 4, -2, -1}$ and evaluating gives $\left( 10, \  - \frac{3888}{125}\right), \left( 4, \  0\right), \left( -2, \  - \frac{432}{125}\right), \left( -1, \  -4\right)$. The max is $\left( 4, \  0\right)$ and the min is $\left( 10, \  - \frac{3888}{125}\right)$.}

\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{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20%5Cfrac%7B8%20x%5E%7B3%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B36%20x%5E%7B2%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B96%20x%7D%7B125%7D%20-%20%5Cfrac%7B448%7D%7B125%7D%20" alt="LaTeX:  \displaystyle f(x) = - \frac{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125} " data-equation-content=" \displaystyle f(x) = - \frac{8 x^{3}}{125} + \frac{36 x^{2}}{125} + \frac{96 x}{125} - \frac{448}{125} " />  on  <img class="equation_image" title=" \displaystyle [-2,10] " src="/equation_images/%20%5Cdisplaystyle%20%5B-2%2C10%5D%20" alt="LaTeX:  \displaystyle [-2,10] " data-equation-content=" \displaystyle [-2,10] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B24%20x%5E%7B2%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B72%20x%7D%7B125%7D%20%2B%20%5Cfrac%7B96%7D%7B125%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} " data-equation-content=" \displaystyle f'(x) = - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B24%20x%5E%7B2%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B72%20x%7D%7B125%7D%20%2B%20%5Cfrac%7B96%7D%7B125%7D%20%3D%200%20" alt="LaTeX:  \displaystyle - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0 " data-equation-content=" \displaystyle - \frac{24 x^{2}}{125} + \frac{72 x}{125} + \frac{96}{125} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1%20" alt="LaTeX:  \displaystyle x = -1 " data-equation-content=" \displaystyle x = -1 " />  and  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " /> . 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 {10, 4, -2, -1} " src="/equation_images/%20%5Cdisplaystyle%20%7B10%2C%204%2C%20-2%2C%20-1%7D%20" alt="LaTeX:  \displaystyle {10, 4, -2, -1} " data-equation-content=" \displaystyle {10, 4, -2, -1} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 10, \  - \frac{3888}{125}\right), \left( 4, \  0\right), \left( -2, \  - \frac{432}{125}\right), \left( -1, \  -4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%2010%2C%20%5C%20%20-%20%5Cfrac%7B3888%7D%7B125%7D%5Cright%29%2C%20%5Cleft%28%204%2C%20%5C%20%200%5Cright%29%2C%20%5Cleft%28%20-2%2C%20%5C%20%20-%20%5Cfrac%7B432%7D%7B125%7D%5Cright%29%2C%20%5Cleft%28%20-1%2C%20%5C%20%20-4%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 10, \  - \frac{3888}{125}\right), \left( 4, \  0\right), \left( -2, \  - \frac{432}{125}\right), \left( -1, \  -4\right) " data-equation-content=" \displaystyle \left( 10, \  - \frac{3888}{125}\right), \left( 4, \  0\right), \left( -2, \  - \frac{432}{125}\right), \left( -1, \  -4\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 4, \  0\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%204%2C%20%5C%20%200%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 4, \  0\right) " data-equation-content=" \displaystyle \left( 4, \  0\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( 10, \  - \frac{3888}{125}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%2010%2C%20%5C%20%20-%20%5Cfrac%7B3888%7D%7B125%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 10, \  - \frac{3888}{125}\right) " data-equation-content=" \displaystyle \left( 10, \  - \frac{3888}{125}\right) " /> .</p> </p>