a. Explanation Definition used: “ Critical points: An interior point of the domain of a function f where f ′ is zero or undefined is a critical point of f .” Procedure used: “ First Derivative test for Local Extrema. Suppose that c is a critical point of a continuous function f , and that f is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right, 1. If f ′ changes from negative to positive at c , then f has a local minimum at c ; 2. If f ′ changes from positive to negative at c , then f has a local maximum at c ; 3. If f ′ does not change sign at c (that is, f ′ is positive on both sides of c or negative on both sides), then f has no local extremum at c .” Calculation: The given function is f ( t ) = t 3 − 3 t 2 in the interval − ∞ < t ≤ 3 . Obtain the local extreme values in the given domain as follows. The first derivative of the function f is, f ′ ( t ) = 3 t 2 − 6 t = 3 t ( t − 2 ) The critical point of the function is obtained as, 3 t ( t − 2 ) = 0 b. To determine The absolute maximum and minimum values of the function f ( t ) = t 3 − 3 t 2 in the interval − ∞ < t ≤ 3 . c. To determine To Sketch: the graph of the function f ( t ) = t 3 − 3 t 2 in the interval − ∞ < t ≤ 3 .

4th Edition

ISBN: 9780134439099

Author: Hass, Joel., Heil, Christopher , WEIR, Maurice D.

Publisher: Pearson,

See similar textbooks

Concept explainers

Transformation of Graphs

The word ‘transformation’ means modification. Transformation of the graph of a function is a process by which we modify or change the original graph and make a new graph.

Exponential Functions

The exponential function is a type of mathematical function which is used in real-world contexts. It helps to find out the exponential decay model or exponential growth model, in mathematical models. In this topic, we will understand descriptive rules, c…

Videos

Question

Chapter 4.3, Problem 46E

To determine

The local maximum and minimum values of the function f(t)=t3−3t2 in the interval −∞<t≤3.

To determine

The absolute maximum and minimum values of the function f(t)=t3−3t2 in the interval −∞<t≤3.

To determine

To Sketch: the graph of the function f(t)=t3−3t2 in the interval −∞<t≤3.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.3, Problem 45E

Chapter 4.3, Problem 47E

Students have asked these similar questions

Evaluate the double integral ' √ √ (−2xy² + 3ry) dA R where R = {(x,y)| 1 ≤ x ≤ 3, 2 ≤ y ≤ 4} Double Integral Plot of integrand and Region R N 120 100 80- 60- 40 20 -20 -40 2 T 3 4 5123456 This plot is an example of the function over region R. The region and function identified in your problem will be slightly different. Answer = Round your answer to four decimal places.

Find Te²+ dydz 0 Write your answer in exact form.

xy² Find -dA, R = [0,3] × [−4,4] x²+1 Round your answer to four decimal places.

Chapter 4 Solutions

Thomas' Calculus

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The local maximum and minimum values of the function f ( t ) = t 3 − 3 t 2 in the interval − ∞ &lt; t ≤ 3 .

Concept explainers

Videos

The local maximum and minimum values of the function f(t)=t3−3t2 in the interval −∞<t≤3.

The absolute maximum and minimum values of the function f(t)=t3−3t2 in the interval −∞<t≤3.

To Sketch: the graph of the function f(t)=t3−3t2 in the interval −∞<t≤3.

Want to see the full answer?

Chapter 4 Solutions

The local maximum and minimum values of the function f ( t ) = t 3 − 3 t 2 in the interval − ∞ < t ≤ 3 .