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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = \frac{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2}\) on \(\displaystyle [-6,6]\)

Taking the derivative gives \(\displaystyle f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 0\). 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 critical number \(\displaystyle x = -6\) is not in the interval \(\displaystyle (-6,6)\) and does not need to be checked. The inputs to be checked are \(\displaystyle {-6, 2, 6}\) and evaluating gives \(\displaystyle \left( -6, \ 10\right), \left( 2, \ -6\right), \left( 6, \ 10\right)\). The graph has a maximum value of \(\displaystyle 10\) at the points \(\displaystyle x = 6,-6\)and the min is \(\displaystyle \left( 2, \ -6\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = \frac{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2}$ on $[-6,6]$
    \soln{9cm}{Taking the derivative gives $f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4}$.  Setting it equal to zero and solving gives the critical numbers. $\frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 0$. 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 critical number $x = -6$ is not in the interval $(-6,6)$ and does not need to be checked. The inputs to be checked are ${-6, 2, 6}$ and evaluating gives $\left( -6, \  10\right), \left( 2, \  -6\right), \left( 6, \  10\right)$. The graph has a maximum value of $10$ at the points $x = 6,-6$and the min is $\left( 2, \  -6\right)$.}

\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{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B3%7D%7D%7B16%7D%20%2B%20%5Cfrac%7B3%20x%5E%7B2%7D%7D%7B8%7D%20-%20%5Cfrac%7B9%20x%7D%7B4%7D%20-%20%5Cfrac%7B7%7D%7B2%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2} " data-equation-content=" \displaystyle f(x) = \frac{x^{3}}{16} + \frac{3 x^{2}}{8} - \frac{9 x}{4} - \frac{7}{2} " />  on  <img class="equation_image" title=" \displaystyle [-6,6] " src="/equation_images/%20%5Cdisplaystyle%20%5B-6%2C6%5D%20" alt="LaTeX:  \displaystyle [-6,6] " data-equation-content=" \displaystyle [-6,6] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B3%20x%5E%7B2%7D%7D%7B16%7D%20%2B%20%5Cfrac%7B3%20x%7D%7B4%7D%20-%20%5Cfrac%7B9%7D%7B4%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} " data-equation-content=" \displaystyle f'(x) = \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B3%20x%5E%7B2%7D%7D%7B16%7D%20%2B%20%5Cfrac%7B3%20x%7D%7B4%7D%20-%20%5Cfrac%7B9%7D%7B4%7D%20%3D%200%20" alt="LaTeX:  \displaystyle \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 0 " data-equation-content=" \displaystyle \frac{3 x^{2}}{16} + \frac{3 x}{4} - \frac{9}{4} = 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 critical number  <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 " />  is not in the interval  <img class="equation_image" title=" \displaystyle (-6,6) " src="/equation_images/%20%5Cdisplaystyle%20%28-6%2C6%29%20" alt="LaTeX:  \displaystyle (-6,6) " data-equation-content=" \displaystyle (-6,6) " />  and does not need to be checked. The inputs to be checked are  <img class="equation_image" title=" \displaystyle {-6, 2, 6} " src="/equation_images/%20%5Cdisplaystyle%20%7B-6%2C%202%2C%206%7D%20" alt="LaTeX:  \displaystyle {-6, 2, 6} " data-equation-content=" \displaystyle {-6, 2, 6} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( -6, \  10\right), \left( 2, \  -6\right), \left( 6, \  10\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-6%2C%20%5C%20%2010%5Cright%29%2C%20%5Cleft%28%202%2C%20%5C%20%20-6%5Cright%29%2C%20%5Cleft%28%206%2C%20%5C%20%2010%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -6, \  10\right), \left( 2, \  -6\right), \left( 6, \  10\right) " data-equation-content=" \displaystyle \left( -6, \  10\right), \left( 2, \  -6\right), \left( 6, \  10\right) " /> . The graph has a maximum value of  <img class="equation_image" title=" \displaystyle 10 " src="/equation_images/%20%5Cdisplaystyle%2010%20" alt="LaTeX:  \displaystyle 10 " data-equation-content=" \displaystyle 10 " />  at the points  <img class="equation_image" title=" \displaystyle x = 6,-6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%206%2C-6%20" alt="LaTeX:  \displaystyle x = 6,-6 " data-equation-content=" \displaystyle x = 6,-6 " /> and the min is  <img class="equation_image" title=" \displaystyle \left( 2, \  -6\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%202%2C%20%5C%20%20-6%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 2, \  -6\right) " data-equation-content=" \displaystyle \left( 2, \  -6\right) " /> .</p> </p>