Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Derivatives / Interpretation of the Derivative
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 3.2 : Interpretation of the Derivative

11. Determine where, if anywhere, the function \(g\left( x \right) = {x^3} - 2{x^2} + x - 1\) stops changing.

Show Solution

We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 7 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.

From our previous work (with a corresponding change of variables) we know that the derivative is,

\[g'\left( x \right) = 3{x^2} - 4x + 1\]

If the function stops changing at a point then the derivative will be zero at that point. So, to determine if we function stops changing we will need to solve,

\[\begin{align*}g'\left( x \right) & = 0\\ 3{x^2} - 4x + 1 & = 0\\ \left( {3x - 1} \right)\left( {x - 1} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{3},\,x = 1\end{align*}\]

So, the function will stop changing at \(x = \frac{1}{3}\) and \(x = 1\).