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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 \(\displaystyle f(x) = 7 x^{3} + 15 x^{2} + 9 x + 2\).

To find the critical numbers solve \(\displaystyle f'(x) = 0\). The derivative is \(\displaystyle f'(x) = 21 x^{2} + 30 x + 9\). Solving \(\displaystyle 21 x^{2} + 30 x + 9 = 0\) gives \(\displaystyle x = \left[ -1, \ - \frac{3}{7}\right]\). Using the 2nd derivative test gives:
\(\displaystyle f''\left( -1 \right) = -12 \) which is less than zero, so the function is concave down and \(\displaystyle f\left(-1\right) = 1\) is a local maximum.
\(\displaystyle f''\left( - \frac{3}{7} \right) = 12 \) which is greater than zero, so the function is concave up and \(\displaystyle f\left(- \frac{3}{7}\right) = \frac{17}{49}\) is a local minimum.

Download \(\LaTeX\) 

\begin{question}Find the local maximum and minimum of $f(x) = 7 x^{3} + 15 x^{2} + 9 x + 2$. 
    \soln{9cm}{To find the critical numbers solve $f'(x) = 0$. The derivative is $f'(x) = 21 x^{2} + 30 x + 9$. Solving $21 x^{2} + 30 x + 9 = 0$ gives $x = \left[ -1, \  - \frac{3}{7}\right]$. Using the 2nd derivative test gives:\newline $f''\left( -1 \right) = -12 $ which is less than zero, so the function is concave down and $f\left(-1\right) = 1$ is a local maximum.\newline $f''\left( - \frac{3}{7} \right) = 12 $ which is greater than zero, so the function is concave up and $f\left(- \frac{3}{7}\right) = \frac{17}{49}$ 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) = 7 x^{3} + 15 x^{2} + 9 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%207%20x%5E%7B3%7D%20%2B%2015%20x%5E%7B2%7D%20%2B%209%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f(x) = 7 x^{3} + 15 x^{2} + 9 x + 2 " data-equation-content=" \displaystyle f(x) = 7 x^{3} + 15 x^{2} + 9 x + 2 " /> . </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} + 30 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%2021%20x%5E%7B2%7D%20%2B%2030%20x%20%2B%209%20" alt="LaTeX:  \displaystyle f'(x) = 21 x^{2} + 30 x + 9 " data-equation-content=" \displaystyle f'(x) = 21 x^{2} + 30 x + 9 " /> . Solving  <img class="equation_image" title=" \displaystyle 21 x^{2} + 30 x + 9 = 0 " src="/equation_images/%20%5Cdisplaystyle%2021%20x%5E%7B2%7D%20%2B%2030%20x%20%2B%209%20%3D%200%20" alt="LaTeX:  \displaystyle 21 x^{2} + 30 x + 9 = 0 " data-equation-content=" \displaystyle 21 x^{2} + 30 x + 9 = 0 " />  gives  <img class="equation_image" title=" \displaystyle x = \left[ -1, \  - \frac{3}{7}\right] " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cleft%5B%20-1%2C%20%5C%20%20-%20%5Cfrac%7B3%7D%7B7%7D%5Cright%5D%20" alt="LaTeX:  \displaystyle x = \left[ -1, \  - \frac{3}{7}\right] " data-equation-content=" \displaystyle x = \left[ -1, \  - \frac{3}{7}\right] " /> . Using the 2nd derivative test gives:<br>  <img class="equation_image" title=" \displaystyle f''\left( -1 \right) = -12  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20-1%20%5Cright%29%20%3D%20-12%20%20" alt="LaTeX:  \displaystyle f''\left( -1 \right) = -12  " data-equation-content=" \displaystyle f''\left( -1 \right) = -12  " />  which is less than zero, so the function is concave down and  <img class="equation_image" title=" \displaystyle f\left(-1\right) = 1 " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28-1%5Cright%29%20%3D%201%20" alt="LaTeX:  \displaystyle f\left(-1\right) = 1 " data-equation-content=" \displaystyle f\left(-1\right) = 1 " />  is a local maximum.<br>  <img class="equation_image" title=" \displaystyle f''\left( - \frac{3}{7} \right) = 12  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20-%20%5Cfrac%7B3%7D%7B7%7D%20%5Cright%29%20%3D%2012%20%20" alt="LaTeX:  \displaystyle f''\left( - \frac{3}{7} \right) = 12  " data-equation-content=" \displaystyle f''\left( - \frac{3}{7} \right) = 12  " />  which is greater than zero, so the function is concave up and  <img class="equation_image" title=" \displaystyle f\left(- \frac{3}{7}\right) = \frac{17}{49} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28-%20%5Cfrac%7B3%7D%7B7%7D%5Cright%29%20%3D%20%5Cfrac%7B17%7D%7B49%7D%20" alt="LaTeX:  \displaystyle f\left(- \frac{3}{7}\right) = \frac{17}{49} " data-equation-content=" \displaystyle f\left(- \frac{3}{7}\right) = \frac{17}{49} " />  is a local minimum.<br> </p> </p>