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check 0 2 and 3 as critical values
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 ?
so you are only suppose to be looking from 0 to 3 and the only critical number you have between there is 2. You test 0 to 2 and also 2 to 3.
You have two intervals to test on [0,3] in other words.
but the function at this point would be increasing all the time , so there will be no absolute max , but I've found the absolute max
so if I took numbers larger than 2 the function would be increasing , and I i found numbers smaller than 2 the function will also be increasing.
and if I took numbers smaller than 2 **
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.
|dw:1386965357606:dw| or you may have something like this |dw:1386965446894:dw| or |dw:1386965478687:dw| or
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 --- 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 ,
was your function continuous on 0 to 3?
i accidentally said what
Ok so you probably have something that make kinda looks like one of the pictures I have above.
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.
maybe possible in a different quadrant
Well why don't you give the polynomial and the endpoints?
Well we have the endpoints
just the polynomial
You will have an abs max and abs min though
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
also local min and local max can't occur at endpoints but abs max and abs min can occur at endpoints
the exam was 3 days ago :(
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.
You can say it is increasing all the time on that interval.
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.
But there is nothing existing outside that interval/
ah man I'm really upset about it :(
There is a lot of words and junk to keep up with in calculus.
Sometimes things can get lost in translation.
@myininaya thank you so much sir.
Sir is close enough.
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
Oh well , I learned from my mistake. Thanks again for helping me , you're the best ^^