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Calculus
Applications of Derivatives
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Find the absolute maximum of \(\displaystyle f(x) = \frac{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125}\) on \(\displaystyle [-9,9]\)

Taking the derivative gives \(\displaystyle f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125}\). Setting it equal to zero and solving gives the critical numbers. \(\displaystyle \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0\). The critical numbers are \(\displaystyle x = -3\) 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 \(\displaystyle {9, 2, -3, -9}\) and evaluating gives \(\displaystyle \left( 9, \ \frac{3013}{125}\right), \left( 2, \ -10\right), \left( -3, \ -7\right), \left( -9, \ - \frac{3791}{125}\right)\). The max is \(\displaystyle \left( 9, \ \frac{3013}{125}\right)\) and the min is \(\displaystyle \left( -9, \ - \frac{3791}{125}\right)\).

Download \(\LaTeX\) 

\begin{question}Find the absolute maximum of $f(x) = \frac{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125}$ on $[-9,9]$
    \soln{9cm}{Taking the derivative gives $f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125}$.  Setting it equal to zero and solving gives the critical numbers. $\frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0$. The critical numbers are $x = -3$ 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 ${9, 2, -3, -9}$ and evaluating gives $\left( 9, \  \frac{3013}{125}\right), \left( 2, \  -10\right), \left( -3, \  -7\right), \left( -9, \  - \frac{3791}{125}\right)$. The max is $\left( 9, \  \frac{3013}{125}\right)$ and the min is $\left( -9, \  - \frac{3791}{125}\right)$.}

\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{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7B6%20x%5E%7B3%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B9%20x%5E%7B2%7D%7D%7B125%7D%20-%20%5Cfrac%7B108%20x%7D%7B125%7D%20-%20%5Cfrac%7B1118%7D%7B125%7D%20" alt="LaTeX:  \displaystyle f(x) = \frac{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125} " data-equation-content=" \displaystyle f(x) = \frac{6 x^{3}}{125} + \frac{9 x^{2}}{125} - \frac{108 x}{125} - \frac{1118}{125} " />  on  <img class="equation_image" title=" \displaystyle [-9,9] " src="/equation_images/%20%5Cdisplaystyle%20%5B-9%2C9%5D%20" alt="LaTeX:  \displaystyle [-9,9] " data-equation-content=" \displaystyle [-9,9] " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B18%20x%5E%7B2%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B18%20x%7D%7B125%7D%20-%20%5Cfrac%7B108%7D%7B125%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} " data-equation-content=" \displaystyle f'(x) = \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} " /> .  Setting it equal to zero and solving gives the critical numbers.  <img class="equation_image" title=" \displaystyle \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B18%20x%5E%7B2%7D%7D%7B125%7D%20%2B%20%5Cfrac%7B18%20x%7D%7B125%7D%20-%20%5Cfrac%7B108%7D%7B125%7D%20%3D%200%20" alt="LaTeX:  \displaystyle \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0 " data-equation-content=" \displaystyle \frac{18 x^{2}}{125} + \frac{18 x}{125} - \frac{108}{125} = 0 " /> . The critical numbers are  <img class="equation_image" title=" \displaystyle x = -3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-3%20" alt="LaTeX:  \displaystyle x = -3 " data-equation-content=" \displaystyle x = -3 " />  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 {9, 2, -3, -9} " src="/equation_images/%20%5Cdisplaystyle%20%7B9%2C%202%2C%20-3%2C%20-9%7D%20" alt="LaTeX:  \displaystyle {9, 2, -3, -9} " data-equation-content=" \displaystyle {9, 2, -3, -9} " />  and evaluating gives  <img class="equation_image" title=" \displaystyle \left( 9, \  \frac{3013}{125}\right), \left( 2, \  -10\right), \left( -3, \  -7\right), \left( -9, \  - \frac{3791}{125}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%209%2C%20%5C%20%20%5Cfrac%7B3013%7D%7B125%7D%5Cright%29%2C%20%5Cleft%28%202%2C%20%5C%20%20-10%5Cright%29%2C%20%5Cleft%28%20-3%2C%20%5C%20%20-7%5Cright%29%2C%20%5Cleft%28%20-9%2C%20%5C%20%20-%20%5Cfrac%7B3791%7D%7B125%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 9, \  \frac{3013}{125}\right), \left( 2, \  -10\right), \left( -3, \  -7\right), \left( -9, \  - \frac{3791}{125}\right) " data-equation-content=" \displaystyle \left( 9, \  \frac{3013}{125}\right), \left( 2, \  -10\right), \left( -3, \  -7\right), \left( -9, \  - \frac{3791}{125}\right) " /> . The max is  <img class="equation_image" title=" \displaystyle \left( 9, \  \frac{3013}{125}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%209%2C%20%5C%20%20%5Cfrac%7B3013%7D%7B125%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( 9, \  \frac{3013}{125}\right) " data-equation-content=" \displaystyle \left( 9, \  \frac{3013}{125}\right) " />  and the min is  <img class="equation_image" title=" \displaystyle \left( -9, \  - \frac{3791}{125}\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-9%2C%20%5C%20%20-%20%5Cfrac%7B3791%7D%7B125%7D%5Cright%29%20" alt="LaTeX:  \displaystyle \left( -9, \  - \frac{3791}{125}\right) " data-equation-content=" \displaystyle \left( -9, \  - \frac{3791}{125}\right) " /> .</p> </p>