Find the following for the function [tex]$p(t)=2 t^3+3 t^2-36 t$[/tex]:
- Second Derivative
- Local max?
- Local min?
- Critical points?
Answer (1)
Find the first derivative: p ′ ( t ) = 6 t 2 + 6 t − 36 .
Find the critical points by solving p ′ ( t ) = 0 , which gives t = − 3 and t = 2 .
Find the second derivative: p ′′ ( t ) = 12 t + 6 .
Use the second derivative test: p ′′ ( − 3 ) = − 30 < 0 (local max) and 0"> p ′′ ( 2 ) = 30 > 0 (local min). The function has a local maximum at t = − 3 and a local minimum at t = 2 .
The critical points are t = − 3 and t = 2 . There is a local max at t = − 3 and a local min at t = 2 .
Local max at t = − 3 , local min at t = 2
Explanation
Problem Setup We are given the function p ( t ) = 2 t 3 + 3 t 2 − 36 t and asked to find its critical points, local maxima, and local minima using the first and second derivatives.
Finding the First Derivative First, we find the first derivative of p ( t ) :
p ′ ( t ) = d t d ( 2 t 3 + 3 t 2 − 36 t ) = 6 t 2 + 6 t − 36
Finding Critical Points Next, we find the critical points by setting p ′ ( t ) = 0 and solving for t :
6 t 2 + 6 t − 36 = 0 t 2 + t − 6 = 0 ( t + 3 ) ( t − 2 ) = 0 Thus, the critical points are t = − 3 and t = 2 .
Finding the Second Derivative Now, we find the second derivative of p ( t ) :
p ′′ ( t ) = d t d ( 6 t 2 + 6 t − 36 ) = 12 t + 6
Second Derivative Test We use the second derivative test to determine whether each critical point is a local maximum or local minimum. We evaluate p ′′ ( t ) at each critical point: For t = − 3 :
p ′′ ( − 3 ) = 12 ( − 3 ) + 6 = − 36 + 6 = − 30 Since p ′′ ( − 3 ) < 0 , there is a local maximum at t = − 3 .
For t = 2 :
p ′′ ( 2 ) = 12 ( 2 ) + 6 = 24 + 6 = 30 Since 0"> p ′′ ( 2 ) > 0 , there is a local minimum at t = 2 .
Conclusion Therefore, the function p ( t ) has a local maximum at t = − 3 and a local minimum at t = 2 .
Examples
Understanding how to find local maxima and minima is crucial in many real-world applications. For example, in business, companies want to maximize profit and minimize costs. By modeling profit or cost as a function and finding its critical points, businesses can determine the optimal production levels or pricing strategies to achieve their goals. Similarly, in physics, finding the minimum potential energy of a system helps determine its stable states.
