An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Find the derivative of the function: f ′ ( x ) = 3 x 2 − 3 x .
Find the critical numbers by setting the derivative to zero: 3 x ( x − 1 ) = 0 , which gives x = 0 and x = 1 .
Evaluate the function at the critical numbers and endpoints: f ( 0 ) = 0 , f ( 1 ) = − 2 1 , f ( − 3 ) = − 40.5 , and f ( 2 ) = 2 .
Determine the absolute minimum and maximum: The absolute minimum is ( − 3 , − 40.5 ) and the absolute maximum is ( 2 , 2 ) . The absolute minimum is ( − 3 , − 40.5 ) and the absolute maximum is ( 2 , 2 ) .
Explanation
Problem Analysis We are given the function f ( x ) = x 3 − 2 3 x 2 and the closed interval [ − 3 , 2 ] . Our goal is to find the absolute extrema of this function on the given interval. This means we need to find the absolute maximum and absolute minimum values of the function on this interval. To do this, we will find the critical points of the function, evaluate the function at these critical points, and evaluate the function at the endpoints of the interval. Then, we will compare all these values to determine the absolute maximum and minimum.
Finding the Derivative First, we need to find the derivative of the function f ( x ) . Using the power rule, we have: f ′ ( x ) = 3 x 2 − 3 x
Finding Critical Numbers Next, we find the critical numbers by setting f ′ ( x ) = 0 and solving for x :
3 x 2 − 3 x = 0 3 x ( x − 1 ) = 0 So, the critical numbers are x = 0 and x = 1 . Both of these critical numbers lie within the open interval ( − 3 , 2 ) .
Evaluating at Critical Numbers Now, we evaluate f ( x ) at each critical number: For x = 0 :
f ( 0 ) = ( 0 ) 3 − 2 3 ( 0 ) 2 = 0 So, the critical point is ( 0 , 0 ) .
For x = 1 :
f ( 1 ) = ( 1 ) 3 − 2 3 ( 1 ) 2 = 1 − 2 3 = − 2 1 So, the critical point is ( 1 , − 2 1 ) .
Evaluating at Endpoints Next, we evaluate f ( x ) at the endpoints of the interval [ − 3 , 2 ] :
For the left endpoint x = − 3 :
f ( − 3 ) = ( − 3 ) 3 − 2 3 ( − 3 ) 2 = − 27 − 2 3 ( 9 ) = − 27 − 2 27 = − 2 54 − 2 27 = − 2 81 = − 40.5 So, the left endpoint is ( − 3 , − 40.5 ) .
For the right endpoint x = 2 :
f ( 2 ) = ( 2 ) 3 − 2 3 ( 2 ) 2 = 8 − 2 3 ( 4 ) = 8 − 6 = 2 So, the right endpoint is ( 2 , 2 ) .
Finding Absolute Extrema Now, we compare the values of f ( x ) at the critical numbers and the endpoints: Critical points: f ( 0 ) = 0 , f ( 1 ) = − 2 1 Endpoints: f ( − 3 ) = − 40.5 , f ( 2 ) = 2 The smallest value is − 40.5 , which occurs at x = − 3 . Therefore, the absolute minimum is ( − 3 , − 40.5 ) .
The largest value is 2 , which occurs at x = 2 . Therefore, the absolute maximum is ( 2 , 2 ) .
Final Answer The critical numbers of f in ( − 3 , 2 ) are x = 0 and x = 1 . Evaluating f at these critical numbers gives the points ( 0 , 0 ) and ( 1 , − 1/2 ) . Evaluating f at the endpoints of [ − 3 , 2 ] gives the points ( − 3 , − 40.5 ) and ( 2 , 2 ) . The absolute minimum of the function on the closed interval [ − 3 , 2 ] is ( − 3 , − 40.5 ) , and the absolute maximum is ( 2 , 2 ) .
Examples
Consider a roller coaster design where the height of the coaster is modeled by the function f ( x ) = x 3 − f r a c 3 2 x 2 over a certain interval representing the track's length. Finding the absolute extrema helps engineers determine the maximum and minimum heights of the roller coaster. This is crucial for safety and design, ensuring the coaster doesn't exceed height restrictions or dip too low, which could affect the ride's performance and safety. Understanding these extrema allows for a thrilling yet safe ride experience.
