Question

find the absolutemaximum and minimum values of each function on the given interval.

Answer 1

\[f(\theta)=\sin (\theta), -\Pi/2\le \theta \le5\Pi/6\]

Answer 2

Well, first off all we need to determine where the orginal function i undefined. In this case it is contious on that given interval. Now we have to obtains the derivaitve of the orginal function and determine where it is zero

Answer 3

Then graph the function. Identify the points on the graph where the absolute extreme occurs, and include their coordinates.

Answer 4

Now the derivaitve of the sinx function i cos(x). Where is this zero?

Answer 5

Specifically within the given interval

Answer 6

the zero for cos?

Answer 7

yes, where is cos(x) zero? In your given interval, it should b zero at , -pi/2 and pi/2

Answer 8

thats what i thought but in the solutions manual it says that -pi/2 is not interior to the domain.

Answer 9

i dont know what they mean by that

Answer 10

but you said, our domain is -pi/2 to 5pi/6 right?

Answer 11

yes

Answer 12

so, now, what you will have to do is evalauate each of the points we have to work with, namely, the two endpoints, -pi/2 and 5pi/6, as well as -pi/2 and pi/2

Answer 13

and you have to evaluate these critical numbers using the orginal function

Answer 14

Now the highest max, you get is your absolute maximum, and the lowest min you get is your absolute minimum. The rest are relative extrema

Answer 15

i know how to do everything but im just trying to figure out why i cant use -pi/2; why it says that it doesnt work because it isnt interioir to the domain

Answer 16

are you sure your boundaries are right, that i to say your interval

Answer 17

perhaps it is -pi/6

Answer 18

yeah im positive, thats why im so confused

Answer 19

hold on

Answer 20

it says it isnt interior to the domain which mean it is part of the domain, but not interior

Answer 21

Right, -pi/2 is not inside the domain, it is simply the endpoint

Answer 22

the problem before this one i did the same thing and i used an endpoint as well though

Answer 23

g(x)=\[\sqrt{4-x ^{2}} -2\le x \le 1\]

Answer 24

the answers should be pi/2 and -pi/2

Answer 25

it lets me use -2 but not 2 because it isnt in the domain

Answer 26

yes, but in out case we can use -pi/2 because it is an endpoint and we have to evaluate the orginal function at that value

Answer 27

g(x)=
so what about in the other one i get a -2 as a critical point but it doesnt tell me i cant use it and its an endpoint as well