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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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Find the local maximum and minimum of \(\displaystyle f(x) = - 7 x^{3} + 33 x^{2} - 9 x + 3\).

To find the critical numbers solve \(\displaystyle f'(x) = 0\). The derivative is \(\displaystyle f'(x) = - 21 x^{2} + 66 x - 9\). Solving \(\displaystyle - 21 x^{2} + 66 x - 9 = 0\) gives \(\displaystyle x = \left[ \frac{1}{7}, \ 3\right]\). Using the 2nd derivative test gives:
\(\displaystyle f''\left( \frac{1}{7} \right) = 60 \) which is greater than zero, so the function is concave up and \(\displaystyle f\left(\frac{1}{7}\right) = \frac{116}{49}\) is a local minimum.
\(\displaystyle f''\left( 3 \right) = -60 \) which is less than zero, so the function is concave down and \(\displaystyle f\left(3\right) = 84\) is a local maximum.

Download \(\LaTeX\) 

\begin{question}Find the local maximum and minimum of $f(x) = - 7 x^{3} + 33 x^{2} - 9 x + 3$. 
    \soln{9cm}{To find the critical numbers solve $f'(x) = 0$. The derivative is $f'(x) = - 21 x^{2} + 66 x - 9$. Solving $- 21 x^{2} + 66 x - 9 = 0$ gives $x = \left[ \frac{1}{7}, \  3\right]$. Using the 2nd derivative test gives:\newline $f''\left( \frac{1}{7} \right) = 60 $ which is greater than zero, so the function is concave up and $f\left(\frac{1}{7}\right) = \frac{116}{49}$ is a local minimum.\newline $f''\left( 3 \right) = -60 $ which is less than zero, so the function is concave down and $f\left(3\right) = 84$ is a local maximum.\newline }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the local maximum and minimum of  <img class="equation_image" title=" \displaystyle f(x) = - 7 x^{3} + 33 x^{2} - 9 x + 3 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%207%20x%5E%7B3%7D%20%2B%2033%20x%5E%7B2%7D%20-%209%20x%20%2B%203%20" alt="LaTeX:  \displaystyle f(x) = - 7 x^{3} + 33 x^{2} - 9 x + 3 " data-equation-content=" \displaystyle f(x) = - 7 x^{3} + 33 x^{2} - 9 x + 3 " /> . </p> </p>
HTML for Canvas
<p> <p>To find the critical numbers solve  <img class="equation_image" title=" \displaystyle f'(x) = 0 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%200%20" alt="LaTeX:  \displaystyle f'(x) = 0 " data-equation-content=" \displaystyle f'(x) = 0 " /> . The derivative is  <img class="equation_image" title=" \displaystyle f'(x) = - 21 x^{2} + 66 x - 9 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%2021%20x%5E%7B2%7D%20%2B%2066%20x%20-%209%20" alt="LaTeX:  \displaystyle f'(x) = - 21 x^{2} + 66 x - 9 " data-equation-content=" \displaystyle f'(x) = - 21 x^{2} + 66 x - 9 " /> . Solving  <img class="equation_image" title=" \displaystyle - 21 x^{2} + 66 x - 9 = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%2021%20x%5E%7B2%7D%20%2B%2066%20x%20-%209%20%3D%200%20" alt="LaTeX:  \displaystyle - 21 x^{2} + 66 x - 9 = 0 " data-equation-content=" \displaystyle - 21 x^{2} + 66 x - 9 = 0 " />  gives  <img class="equation_image" title=" \displaystyle x = \left[ \frac{1}{7}, \  3\right] " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cleft%5B%20%5Cfrac%7B1%7D%7B7%7D%2C%20%5C%20%203%5Cright%5D%20" alt="LaTeX:  \displaystyle x = \left[ \frac{1}{7}, \  3\right] " data-equation-content=" \displaystyle x = \left[ \frac{1}{7}, \  3\right] " /> . Using the 2nd derivative test gives:<br>  <img class="equation_image" title=" \displaystyle f''\left( \frac{1}{7} \right) = 60  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20%5Cfrac%7B1%7D%7B7%7D%20%5Cright%29%20%3D%2060%20%20" alt="LaTeX:  \displaystyle f''\left( \frac{1}{7} \right) = 60  " data-equation-content=" \displaystyle f''\left( \frac{1}{7} \right) = 60  " />  which is greater than zero, so the function is concave up and  <img class="equation_image" title=" \displaystyle f\left(\frac{1}{7}\right) = \frac{116}{49} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28%5Cfrac%7B1%7D%7B7%7D%5Cright%29%20%3D%20%5Cfrac%7B116%7D%7B49%7D%20" alt="LaTeX:  \displaystyle f\left(\frac{1}{7}\right) = \frac{116}{49} " data-equation-content=" \displaystyle f\left(\frac{1}{7}\right) = \frac{116}{49} " />  is a local minimum.<br>  <img class="equation_image" title=" \displaystyle f''\left( 3 \right) = -60  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%203%20%5Cright%29%20%3D%20-60%20%20" alt="LaTeX:  \displaystyle f''\left( 3 \right) = -60  " data-equation-content=" \displaystyle f''\left( 3 \right) = -60  " />  which is less than zero, so the function is concave down and  <img class="equation_image" title=" \displaystyle f\left(3\right) = 84 " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%283%5Cright%29%20%3D%2084%20" alt="LaTeX:  \displaystyle f\left(3\right) = 84 " data-equation-content=" \displaystyle f\left(3\right) = 84 " />  is a local maximum.<br> </p> </p>