I messed up :(
suppose we have a function on the interval [0,3]
and we want to determine where the function is increasing and decreasing , and our critical numbers were 2 and 4
what I did it since 4 is not on the interval i put 0 and 3 and took numbers greater than 3 and smaller than 0 and between 0 and 3 :(
what is the correct way ?
should I put only 2 ?
if I did , the function would be increasing all the time , so there will be no absolute max , but there was an absolute max and min.
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I messed up :(
suppose we have a function on the interval [0,3]
and we want to determine where the function is increasing and decreasing , and our critical numbers were 2 and 4
what I did it since 4 is not on the interval i put 0 and 3 and took numbers greater than 3 and smaller than 0 and between 0 and 3 :(
what is the correct way ?
should I put only 2 ?
if I did , the function would be increasing all the time , so there will be no absolute max , but there was an absolute max and min.
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I know , the question was on two parts , part one asks for the absolute max,mid , and there was a large value and a small value , so the largest was the max and the smallest was the min,
but on part 2 , he asks for where the functions increasing and decreasing ,
if I put 2 the function would be increasing so there will be no absolute max , what should I put on the numbers line ?
You will have absolute max at the highest point
You will have absolute min at the lowest point
There will be a highest and lowest point since you are only looking from 0 to 3 (we know our function will not be going on forever)
what I've learned is when the function get to negative when we take numbers smaller or greater the function is decreasing , and if we took numbers smaller or greater and it we got a positive answer , the function is then increasing.
the function when I took 1 were positive , and when I took 2.5 and 3 it was also positive.
so it doesn't matter if it's positive or not ?
our prof said that when the function get to negative when you evaluate a number smaller than the critical number in this case 2 , then it's decreasing , and it's the opposite when it's positive.
To test if the function increasing or decreasing find f'
Plug x values into f' to see if it increasing or decreasing.
positive numbers imply positive slopes which imply increasing
negative numbers imply negative slopes which imply decreasing
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What your function continuous from 0 to 3? In my pictures I assumed it was.
not sure about what you mean here " What your function continuous " but it's polynomial so it's continuous everywhere.
as you said , I plugged 1 and it was positive , and I plugged 2.5 and 3 and it was also positive ,
I'll check it soon ,
:( I just wanted to check that should I put 0 and 3 on the numbers line , I think my answer was wrong..
oh well , thank you anyways.
I don't remember it , but I remember that I got a critical numbers of x=2 and x=4 since 4 is not on [0,3] I plugged 0 and 3 and 2 and I got on one of them 21 , on the other 19 and 1 on 0 I think.
so the absolute max =21 and min = 1
If you are just looking for abs min or abs max on a specific domain restriction with the function being continuous and differentiateable which polynomials are all you have to do
is find the derivative of y=f(x) , [0,3]
y'=f'(x)
Set f'=0 and solve for x to get critical numbers
Say we find the critical numbers x=2 and x=4
(x=4 is something we don't have to worry about since it isn't between 0 and 3)
so we have the number to plug in are x=0,2,3
plug them into y=f(x)
f(0)=
f(2)=
f(3)=
Which ever one of those outputs is biggest, is your absolute max value
Which ever one of those outputs is smallest, is your absolute min value
what got me wrong is a little thing I heard on the last lecture before the exam " when you want to check if there's an absolute max and absolute min , if it's increasing all the time then there's no absolute max " I first put 2 and did what you said , and it was increasing all the time , then crossed it " stupid me :( " and put 0 and 3 and took numbers smaller than 0 and greater than 3 and between 0 and 3
If you want to find where y=f(x) with domain restriction (being differentiateable and continuous) [0,3] is increasing or decreasing
Find y'=f'(x)
Set f'=0 to find critical numbers say you get those numbers are x=2 and =4
ignore x=4 because it is not in the domain of [0,3]
So we have the interval from 0 to 2 to test
and also the interval from 2 to 3 to test
You plug numbers in from those intervals into f' to see if the function is increasing or decreasing
You only have two intervals so you only have two numbers you need to plug in
f'(1) and f'(2.5)
If f'(1) is negative then the function is decreasing on (0,2).
If f'(1) is positive then the function is increasing on (0,2).
If f'(2.5) is negative then the function is decreasing on (2,3).
If f'(2.5) is positive then the function is increasing on (2,3).
It isn't increasing all the the time if you have a function that is continuous and differentiable on a restricted domain. The function will like totally have endpoints if you are given the function that is continuous and differentiateable with a restricted domain.
I did what you said , I took 1 and it was positive , and I took 2.5 and it was also positive .
anyways , thank you so much for helping me , I'll check the question on the next class.
Just so you know you would have been correct if you didn't have the domain restriction and you found the polynomial was increasing all the time.
Like there would be no absolute min or absolute max.
Something if it was decreasing all the time.
Like f(x)=x^3 is good example
f(x)=x^3 increases the whole time so there is no max or min