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Calculus
Applications of Derivatives
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Find the local maximum and minimum of f(x)=−x3+5x2−7x+2.

To find the critical numbers solve f′(x)=0. The derivative is f′(x)=−3x2+10x−7. Solving −3x2+10x−7=0 gives x=[1, 73]. Using the 2nd derivative test gives:
f″(1)=4 which is greater than zero, so the function is concave up and f(1)=−1 is a local minimum.
f″(73)=−4 which is less than zero, so the function is concave down and f(73)=527 is a local maximum.

Download LATEX 

\begin{question}Find the local maximum and minimum of $f(x) = - x^{3} + 5 x^{2} - 7 x + 2$. 
    \soln{9cm}{To find the critical numbers solve $f'(x) = 0$. The derivative is $f'(x) = - 3 x^{2} + 10 x - 7$. Solving $- 3 x^{2} + 10 x - 7 = 0$ gives $x = \left[ 1, \  \frac{7}{3}\right]$. Using the 2nd derivative test gives:\newline $f''\left( 1 \right) = 4 $ which is greater than zero, so the function is concave up and $f\left(1\right) = -1$ is a local minimum.\newline $f''\left( \frac{7}{3} \right) = -4 $ which is less than zero, so the function is concave down and $f\left(\frac{7}{3}\right) = \frac{5}{27}$ is a local maximum.\newline }

\end{question}

Download Question and Solution EnvironmentLATEX
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HTML for Canvas 
<p> <p>Find the local maximum and minimum of  <img class="equation_image" title=" \displaystyle f(x) = - x^{3} + 5 x^{2} - 7 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20-%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20-%207%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f(x) = - x^{3} + 5 x^{2} - 7 x + 2 " data-equation-content=" \displaystyle f(x) = - x^{3} + 5 x^{2} - 7 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) = - 3 x^{2} + 10 x - 7 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%203%20x%5E%7B2%7D%20%2B%2010%20x%20-%207%20" alt="LaTeX:  \displaystyle f'(x) = - 3 x^{2} + 10 x - 7 " data-equation-content=" \displaystyle f'(x) = - 3 x^{2} + 10 x - 7 " /> . Solving  <img class="equation_image" title=" \displaystyle - 3 x^{2} + 10 x - 7 = 0 " src="/equation_images/%20%5Cdisplaystyle%20-%203%20x%5E%7B2%7D%20%2B%2010%20x%20-%207%20%3D%200%20" alt="LaTeX:  \displaystyle - 3 x^{2} + 10 x - 7 = 0 " data-equation-content=" \displaystyle - 3 x^{2} + 10 x - 7 = 0 " />  gives  <img class="equation_image" title=" \displaystyle x = \left[ 1, \  \frac{7}{3}\right] " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cleft%5B%201%2C%20%5C%20%20%5Cfrac%7B7%7D%7B3%7D%5Cright%5D%20" alt="LaTeX:  \displaystyle x = \left[ 1, \  \frac{7}{3}\right] " data-equation-content=" \displaystyle x = \left[ 1, \  \frac{7}{3}\right] " /> . Using the 2nd derivative test gives:<br>  <img class="equation_image" title=" \displaystyle f''\left( 1 \right) = 4  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%201%20%5Cright%29%20%3D%204%20%20" alt="LaTeX:  \displaystyle f''\left( 1 \right) = 4  " data-equation-content=" \displaystyle f''\left( 1 \right) = 4  " />  which is greater than zero, so the function is concave up and  <img class="equation_image" title=" \displaystyle f\left(1\right) = -1 " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%281%5Cright%29%20%3D%20-1%20" alt="LaTeX:  \displaystyle f\left(1\right) = -1 " data-equation-content=" \displaystyle f\left(1\right) = -1 " />  is a local minimum.<br>  <img class="equation_image" title=" \displaystyle f''\left( \frac{7}{3} \right) = -4  " src="/equation_images/%20%5Cdisplaystyle%20f%27%27%5Cleft%28%20%5Cfrac%7B7%7D%7B3%7D%20%5Cright%29%20%3D%20-4%20%20" alt="LaTeX:  \displaystyle f''\left( \frac{7}{3} \right) = -4  " data-equation-content=" \displaystyle f''\left( \frac{7}{3} \right) = -4  " />  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{5}{27} " src="/equation_images/%20%5Cdisplaystyle%20f%5Cleft%28%5Cfrac%7B7%7D%7B3%7D%5Cright%29%20%3D%20%5Cfrac%7B5%7D%7B27%7D%20" alt="LaTeX:  \displaystyle f\left(\frac{7}{3}\right) = \frac{5}{27} " data-equation-content=" \displaystyle f\left(\frac{7}{3}\right) = \frac{5}{27} " />  is a local maximum.<br> </p> </p>