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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 \(\displaystyle f(x) = 3 x^{3} + 12 x^{2} + 7 x - 4\).

To find the critical numbers solve \(\displaystyle f'(x) = 0\). The derivative is \(\displaystyle f'(x) = 9 x^{2} + 24 x + 7\). Solving \(\displaystyle 9 x^{2} + 24 x + 7 = 0\) gives \(\displaystyle x = \left[ - \frac{7}{3}, \ - \frac{1}{3}\right]\). Using the 2nd derivative test gives:
\(\displaystyle f''\left( - \frac{7}{3} \right) = -18 \) which is less than zero, so the function is concave down and \(\displaystyle f\left(- \frac{7}{3}\right) = \frac{62}{9}\) is a local maximum.
\(\displaystyle f''\left( - \frac{1}{3} \right) = 18 \) which is greater than zero, so the function is concave up and \(\displaystyle f\left(- \frac{1}{3}\right) = - \frac{46}{9}\) is a local minimum.

Download \(\LaTeX\) 

\begin{question}Find the local maximum and minimum of $f(x) = 3 x^{3} + 12 x^{2} + 7 x - 4$. 
    \soln{9cm}{To find the critical numbers solve $f'(x) = 0$. The derivative is $f'(x) = 9 x^{2} + 24 x + 7$. Solving $9 x^{2} + 24 x + 7 = 0$ gives $x = \left[ - \frac{7}{3}, \  - \frac{1}{3}\right]$. Using the 2nd derivative test gives:\newline $f''\left( - \frac{7}{3} \right) = -18 $ which is less than zero, so the function is concave down and $f\left(- \frac{7}{3}\right) = \frac{62}{9}$ is a local maximum.\newline $f''\left( - \frac{1}{3} \right) = 18 $ which is greater than zero, so the function is concave up and $f\left(- \frac{1}{3}\right) = - \frac{46}{9}$ is a local minimum.\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) = 3 x^{3} + 12 x^{2} + 7 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%203%20x%5E%7B3%7D%20%2B%2012%20x%5E%7B2%7D%20%2B%207%20x%20-%204%20" alt="LaTeX:  \displaystyle f(x) = 3 x^{3} + 12 x^{2} + 7 x - 4 " data-equation-content=" \displaystyle f(x) = 3 x^{3} + 12 x^{2} + 7 x - 4 " /> . </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) = 9 x^{2} + 24 x + 7 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%209%20x%5E%7B2%7D%20%2B%2024%20x%20%2B%207%20" alt="LaTeX:  \displaystyle f'(x) = 9 x^{2} + 24 x + 7 " data-equation-content=" \displaystyle f'(x) = 9 x^{2} + 24 x + 7 " /> . Solving  <img class="equation_image" title=" \displaystyle 9 x^{2} + 24 x + 7 = 0 " src="/equation_images/%20%5Cdisplaystyle%209%20x%5E%7B2%7D%20%2B%2024%20x%20%2B%207%20%3D%200%20" alt="LaTeX:  \displaystyle 9 x^{2} + 24 x + 7 = 0 " data-equation-content=" \displaystyle 9 x^{2} + 24 x + 7 = 0 " />  gives  <img class="equation_image" title=" \displaystyle x = \left[ - \frac{7}{3}, \  - \frac{1}{3}\right] " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cleft%5B%20-%20%5Cfrac%7B7%7D%7B3%7D%2C%20%5C%20%20-%20%5Cfrac%7B1%7D%7B3%7D%5Cright%5D%20" alt="LaTeX:  \displaystyle x = \left[ - \frac{7}{3}, \  - \frac{1}{3}\right] " data-equation-content=" \displaystyle x = \left[ - \frac{7}{3}, \  - \frac{1}{3}\right] " /> . Using the 2nd derivative test gives:<br>  <img class="equation_image" title=" \displaystyle f''\left( - \frac{7}{3} \right) = -18  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20-%20%5Cfrac%7B7%7D%7B3%7D%20%5Cright%29%20%3D%20-18%20%20" alt="LaTeX:  \displaystyle f''\left( - \frac{7}{3} \right) = -18  " data-equation-content=" \displaystyle f''\left( - \frac{7}{3} \right) = -18  " />  which is less than zero, so the function is concave down and  <img class="equation_image" title=" \displaystyle f\left(- \frac{7}{3}\right) = \frac{62}{9} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28-%20%5Cfrac%7B7%7D%7B3%7D%5Cright%29%20%3D%20%5Cfrac%7B62%7D%7B9%7D%20" alt="LaTeX:  \displaystyle f\left(- \frac{7}{3}\right) = \frac{62}{9} " data-equation-content=" \displaystyle f\left(- \frac{7}{3}\right) = \frac{62}{9} " />  is a local maximum.<br>  <img class="equation_image" title=" \displaystyle f''\left( - \frac{1}{3} \right) = 18  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20-%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%3D%2018%20%20" alt="LaTeX:  \displaystyle f''\left( - \frac{1}{3} \right) = 18  " data-equation-content=" \displaystyle f''\left( - \frac{1}{3} \right) = 18  " />  which is greater than zero, so the function is concave up and  <img class="equation_image" title=" \displaystyle f\left(- \frac{1}{3}\right) = - \frac{46}{9} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28-%20%5Cfrac%7B1%7D%7B3%7D%5Cright%29%20%3D%20-%20%5Cfrac%7B46%7D%7B9%7D%20" alt="LaTeX:  \displaystyle f\left(- \frac{1}{3}\right) = - \frac{46}{9} " data-equation-content=" \displaystyle f\left(- \frac{1}{3}\right) = - \frac{46}{9} " />  is a local minimum.<br> </p> </p>