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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = - \frac{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32}\) on \(\displaystyle [-9,5]\)

Taking the derivative gives \(\displaystyle f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 0\). The critical numbers are \(\displaystyle x = -1\) and \(\displaystyle x = 3\). The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are \(\displaystyle {3, 5, -1, -9}\) and evaluating gives \(\displaystyle \left( 3, \ 9\right), \left( 5, \ -2\right), \left( -1, \ -2\right), \left( -9, \ 306\right)\). The max is \(\displaystyle \left( -9, \ 306\right)\) and the graph has a minimum value of \(\displaystyle -2\) at the points \(\displaystyle x = -1,5\)

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = - \frac{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32}$ on $[-9,5]$
    \soln{9cm}{Taking the derivative gives $f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32}$.  Setting it equal to zero and solving gives the critical numbers. $- \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 0$. The critical numbers are $x = -1$ and $x = 3$. The absolute maximum is either at a critical number or at the end point of the interval. The inputs to be checked are ${3, 5, -1, -9}$ and evaluating gives $\left( 3, \  9\right), \left( 5, \  -2\right), \left( -1, \  -2\right), \left( -9, \  306\right)$. The max is $\left( -9, \  306\right)$ and the graph has a minimum value of $-2$ at the points $x = -1,5$}

\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{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20%5Cfrac%7B11%20x%5E%7B3%7D%7D%7B32%7D%20%2B%20%5Cfrac%7B33%20x%5E%7B2%7D%7D%7B32%7D%20%2B%20%5Cfrac%7B99%20x%7D%7B32%7D%20-%20%5Cfrac%7B9%7D%7B32%7D%20" alt="LaTeX:  \displaystyle f(x) = - \frac{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32} " data-equation-content=" \displaystyle f(x) = - \frac{11 x^{3}}{32} + \frac{33 x^{2}}{32} + \frac{99 x}{32} - \frac{9}{32} " />  on  <img class="equation_image" title=" \displaystyle [-9,5] " src="/equation_images/%20%5Cdisplaystyle%20%5B-9%2C5%5D%20" alt="LaTeX:  \displaystyle [-9,5] " data-equation-content=" \displaystyle [-9,5] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B33%20x%5E%7B2%7D%7D%7B32%7D%20%2B%20%5Cfrac%7B33%20x%7D%7B16%7D%20%2B%20%5Cfrac%7B99%7D%7B32%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} " data-equation-content=" \displaystyle f'(x) = - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B33%20x%5E%7B2%7D%7D%7B32%7D%20%2B%20%5Cfrac%7B33%20x%7D%7B16%7D%20%2B%20%5Cfrac%7B99%7D%7B32%7D%20%3D%200%20" alt="LaTeX:  \displaystyle - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 0 " data-equation-content=" \displaystyle - \frac{33 x^{2}}{32} + \frac{33 x}{16} + \frac{99}{32} = 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 = 3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%203%20" alt="LaTeX:  \displaystyle x = 3 " data-equation-content=" \displaystyle x = 3 " /> . 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 {3, 5, -1, -9} " src="/equation_images/%20%5Cdisplaystyle%20%7B3%2C%205%2C%20-1%2C%20-9%7D%20" alt="LaTeX:  \displaystyle {3, 5, -1, -9} " data-equation-content=" \displaystyle {3, 5, -1, -9} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 3, \  9\right), \left( 5, \  -2\right), \left( -1, \  -2\right), \left( -9, \  306\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%203%2C%20%5C%20%209%5Cright%29%2C%20%5Cleft%28%205%2C%20%5C%20%20-2%5Cright%29%2C%20%5Cleft%28%20-1%2C%20%5C%20%20-2%5Cright%29%2C%20%5Cleft%28%20-9%2C%20%5C%20%20306%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 3, \  9\right), \left( 5, \  -2\right), \left( -1, \  -2\right), \left( -9, \  306\right) " data-equation-content=" \displaystyle \left( 3, \  9\right), \left( 5, \  -2\right), \left( -1, \  -2\right), \left( -9, \  306\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( -9, \  306\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-9%2C%20%5C%20%20306%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -9, \  306\right) " data-equation-content=" \displaystyle \left( -9, \  306\right) " />  and the graph has a minimum value of  <img class="equation_image" title=" \displaystyle -2 " src="/equation_images/%20%5Cdisplaystyle%20-2%20" alt="LaTeX:  \displaystyle -2 " data-equation-content=" \displaystyle -2 " />  at the points  <img class="equation_image" title=" \displaystyle x = -1,5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1%2C5%20" alt="LaTeX:  \displaystyle x = -1,5 " data-equation-content=" \displaystyle x = -1,5 " /> </p> </p>