Following on from our discussion from class, complete the following question.
Let \(f(x)=\frac{1}{3}x^3-2x^2\). Find the coordinates of the local maximum and the local minimum of \(f\).
Are you interested in seeing something else that’s sort of neat? Read on.
We’ll eventually be discussing something called the second derivative. Once you’ve found the derivative of a given function, you can then go on to find the derivative of that derivative. For example, if \(f(x)=2x^5\), then \(f'(x)=10x^4\). The second derivative is represented as \(f^{\prime \prime}(x)\); in this case, we have \(f^{\prime \prime}(x)=40x^3\).
Find the second derivative of the function \(f(x)=\frac{1}{3}x^3-2x^2\), then solve the equation \(f^{\prime \prime}(x)=0\). On the graph of \(f\), plot the point on \(f\) whose \(x\)-coordinate is the solution you found to \(f^{\prime \prime}(x)=0\). What do you notice about the location of this point?