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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = - \frac{7 x^{3}}{256} - \frac{21 x^{2}}{128} + \frac{63 x}{64} - \frac{3}{32}\) on \(\displaystyle [-7,8]\)

Taking the derivative gives \(\displaystyle f'(x) = - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} = 0\). The critical numbers are \(\displaystyle x = -6\) 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 \(\displaystyle {8, -7, -6, 2}\) and evaluating gives \(\displaystyle \left( 8, \ - \frac{535}{32}\right), \left( -7, \ - \frac{1445}{256}\right), \left( -6, \ -6\right), \left( 2, \ 1\right)\). The max is \(\displaystyle \left( 2, \ 1\right)\) and the min is \(\displaystyle \left( 8, \ - \frac{535}{32}\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = - \frac{7 x^{3}}{256} - \frac{21 x^{2}}{128} + \frac{63 x}{64} - \frac{3}{32}$ on $[-7,8]$
    \soln{9cm}{Taking the derivative gives $f'(x) = - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64}$.  Setting it equal to zero and solving gives the critical numbers. $- \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} = 0$. The critical numbers are $x = -6$ 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 ${8, -7, -6, 2}$ and evaluating gives $\left( 8, \  - \frac{535}{32}\right), \left( -7, \  - \frac{1445}{256}\right), \left( -6, \  -6\right), \left( 2, \  1\right)$. The max is $\left( 2, \  1\right)$ and the min is $\left( 8, \  - \frac{535}{32}\right)$.}

\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{7 x^{3}}{256} - \frac{21 x^{2}}{128} + \frac{63 x}{64} - \frac{3}{32} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20%5Cfrac%7B7%20x%5E%7B3%7D%7D%7B256%7D%20-%20%5Cfrac%7B21%20x%5E%7B2%7D%7D%7B128%7D%20%2B%20%5Cfrac%7B63%20x%7D%7B64%7D%20-%20%5Cfrac%7B3%7D%7B32%7D%20" alt="LaTeX:  \displaystyle f(x) = - \frac{7 x^{3}}{256} - \frac{21 x^{2}}{128} + \frac{63 x}{64} - \frac{3}{32} " data-equation-content=" \displaystyle f(x) = - \frac{7 x^{3}}{256} - \frac{21 x^{2}}{128} + \frac{63 x}{64} - \frac{3}{32} " />  on  <img class="equation_image" title=" \displaystyle [-7,8] " src="/equation_images/%20%5Cdisplaystyle%20%5B-7%2C8%5D%20" alt="LaTeX:  \displaystyle [-7,8] " data-equation-content=" \displaystyle [-7,8] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B21%20x%5E%7B2%7D%7D%7B256%7D%20-%20%5Cfrac%7B21%20x%7D%7B64%7D%20%2B%20%5Cfrac%7B63%7D%7B64%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} " data-equation-content=" \displaystyle f'(x) = - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B21%20x%5E%7B2%7D%7D%7B256%7D%20-%20%5Cfrac%7B21%20x%7D%7B64%7D%20%2B%20%5Cfrac%7B63%7D%7B64%7D%20%3D%200%20" alt="LaTeX:  \displaystyle - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} = 0 " data-equation-content=" \displaystyle - \frac{21 x^{2}}{256} - \frac{21 x}{64} + \frac{63}{64} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-6%20" alt="LaTeX:  \displaystyle x = -6 " data-equation-content=" \displaystyle x = -6 " />  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 {8, -7, -6, 2} " src="/equation_images/%20%5Cdisplaystyle%20%7B8%2C%20-7%2C%20-6%2C%202%7D%20" alt="LaTeX:  \displaystyle {8, -7, -6, 2} " data-equation-content=" \displaystyle {8, -7, -6, 2} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 8, \  - \frac{535}{32}\right), \left( -7, \  - \frac{1445}{256}\right), \left( -6, \  -6\right), \left( 2, \  1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%208%2C%20%5C%20%20-%20%5Cfrac%7B535%7D%7B32%7D%5Cright%29%2C%20%5Cleft%28%20-7%2C%20%5C%20%20-%20%5Cfrac%7B1445%7D%7B256%7D%5Cright%29%2C%20%5Cleft%28%20-6%2C%20%5C%20%20-6%5Cright%29%2C%20%5Cleft%28%202%2C%20%5C%20%201%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 8, \  - \frac{535}{32}\right), \left( -7, \  - \frac{1445}{256}\right), \left( -6, \  -6\right), \left( 2, \  1\right) " data-equation-content=" \displaystyle \left( 8, \  - \frac{535}{32}\right), \left( -7, \  - \frac{1445}{256}\right), \left( -6, \  -6\right), \left( 2, \  1\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 2, \  1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%201%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  1\right) " data-equation-content=" \displaystyle \left( 2, \  1\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( 8, \  - \frac{535}{32}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%208%2C%20%5C%20%20-%20%5Cfrac%7B535%7D%7B32%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 8, \  - \frac{535}{32}\right) " data-equation-content=" \displaystyle \left( 8, \  - \frac{535}{32}\right) " /> .</p> </p>