Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses

Georgia Southern University

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 46

a. Identify the function's local extreme values in the given domain, and say where they occur.

b. Which of the extreme values, if any, are absolute?

c. Support your findings with a graphing calculator or computer grapher.

$$

f(t)=t^{3}-3 t^{2}, \quad-\infty< t \leq 3

$$

Answer

$(a)[\text { local maximum at } t=0)(b)[\text { absolute maximum at } t=0,3$

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

Okay, So you have. Ah, a tea. It's tick you to minus three. T squared. Santee, between your infinity hand three says, take the derivative to find our critical points. We get three t squared minus sixteen instead of that crime zero. We have that, uh, in fact, two out of three T three tee times T minus two zero. So either t zero tea is too stroller number line. You have zero and two, and then the end of our domain, which is three. So it's like a deaf prime Texas less than zero. This is negative and negatives of positive if, uh, f promise between your into this is negative, but this is positive and negative. And if x is greater than to have positive. So our function is going like that and stopping at three. So we see that we have a local max where well, at zero zero and three zero be plugging three, begin zero zero zero, and then we have a local men, and okay, Tio, what do we get? Money plucking too. Eight minus twelve. So negative for there's not going to be an absolute men because the functions coming from minus infinity. But the absolute max is going to be zero and occurs in X equals zero and X equals three

## Recommended Questions