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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 4.2 : Critical Points
2. Determine the critical points of \(R\left( t \right) = 1 + 80{t^3} + 5{t^4} - 2{t^5}\).
Show All Steps Hide All Steps
Start SolutionWe’ll need the first derivative to get the answer to this problem so let’s get that.
\[R'\left( t \right) = 240{t^2} + 20{t^3} - 10{t^4} = - 10{t^2}\left( {t + 4} \right)\left( {t - 6} \right)\]Factoring the derivative as much as possible will help with the next step.
Show Step 2Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points.
\[ - 10{t^2}\left( {t + 4} \right)\left( {t - 6} \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,\,\,t = - 4,\,\,\,\,t = 6}}\]