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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = \frac{8 x^{3}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9}\) on \(\displaystyle [-4,4]\)

Taking the derivative gives \(\displaystyle f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 0\). The critical numbers are \(\displaystyle x = -1\) 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 \(\displaystyle {2, -4, 4, -1}\) and evaluating gives \(\displaystyle \left( 2, \ -2\right), \left( -4, \ -50\right), \left( 4, \ \frac{190}{9}\right), \left( -1, \ 10\right)\). The max is \(\displaystyle \left( 4, \ \frac{190}{9}\right)\) and the min is \(\displaystyle \left( -4, \ -50\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = \frac{8 x^{3}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9}$ on $[-4,4]$
    \soln{9cm}{Taking the derivative gives $f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3}$.  Setting it equal to zero and solving gives the critical numbers. $\frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 0$. The critical numbers are $x = -1$ 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 ${2, -4, 4, -1}$ and evaluating gives $\left( 2, \  -2\right), \left( -4, \  -50\right), \left( 4, \  \frac{190}{9}\right), \left( -1, \  10\right)$. The max is $\left( 4, \  \frac{190}{9}\right)$ and the min is $\left( -4, \  -50\right)$.}

\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}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7B8%20x%5E%7B3%7D%7D%7B9%7D%20-%20%5Cfrac%7B4%20x%5E%7B2%7D%7D%7B3%7D%20-%20%5Cfrac%7B16%20x%7D%7B3%7D%20%2B%20%5Cfrac%7B62%7D%7B9%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{8 x^{3}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9} " data-equation-content=" \displaystyle f(x) = \frac{8 x^{3}}{9} - \frac{4 x^{2}}{3} - \frac{16 x}{3} + \frac{62}{9} " />  on  <img class="equation_image" title=" \displaystyle [-4,4] " src="/equation_images/%20%5Cdisplaystyle%20%5B-4%2C4%5D%20" alt="LaTeX:  \displaystyle [-4,4] " data-equation-content=" \displaystyle [-4,4] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B8%20x%5E%7B2%7D%7D%7B3%7D%20-%20%5Cfrac%7B8%20x%7D%7B3%7D%20-%20%5Cfrac%7B16%7D%7B3%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} " data-equation-content=" \displaystyle f'(x) = \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B8%20x%5E%7B2%7D%7D%7B3%7D%20-%20%5Cfrac%7B8%20x%7D%7B3%7D%20-%20%5Cfrac%7B16%7D%7B3%7D%20%3D%200%20" alt="LaTeX:  \displaystyle \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 0 " data-equation-content=" \displaystyle \frac{8 x^{2}}{3} - \frac{8 x}{3} - \frac{16}{3} = 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 = 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 {2, -4, 4, -1} " src="/equation_images/%20%5Cdisplaystyle%20%7B2%2C%20-4%2C%204%2C%20-1%7D%20" alt="LaTeX:  \displaystyle {2, -4, 4, -1} " data-equation-content=" \displaystyle {2, -4, 4, -1} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 2, \  -2\right), \left( -4, \  -50\right), \left( 4, \  \frac{190}{9}\right), \left( -1, \  10\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%20-2%5Cright%29%2C%20%5Cleft%28%20-4%2C%20%5C%20%20-50%5Cright%29%2C%20%5Cleft%28%204%2C%20%5C%20%20%5Cfrac%7B190%7D%7B9%7D%5Cright%29%2C%20%5Cleft%28%20-1%2C%20%5C%20%2010%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  -2\right), \left( -4, \  -50\right), \left( 4, \  \frac{190}{9}\right), \left( -1, \  10\right) " data-equation-content=" \displaystyle \left( 2, \  -2\right), \left( -4, \  -50\right), \left( 4, \  \frac{190}{9}\right), \left( -1, \  10\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 4, \  \frac{190}{9}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%204%2C%20%5C%20%20%5Cfrac%7B190%7D%7B9%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 4, \  \frac{190}{9}\right) " data-equation-content=" \displaystyle \left( 4, \  \frac{190}{9}\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( -4, \  -50\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-4%2C%20%5C%20%20-50%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -4, \  -50\right) " data-equation-content=" \displaystyle \left( -4, \  -50\right) " /> .</p> </p>