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anonymous
 one year ago
if, for all x, f'(x)=(x2)^4(x1)^3 , it follows that the function f has
anonymous
 one year ago
if, for all x, f'(x)=(x2)^4(x1)^3 , it follows that the function f has

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so you need help integrating then \[\int\limits (x2)^2(x1)^3 dx \] Let u=x1 then du=dx if u=x1 subtracting one on both sides gives: u1=x2 so you have \[\int\limits_{}^{}(u1)^2u^3 du\] this is just to make the multiplication less harsh we still have to multiply

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so multiply out (u1)^2 then multiply each one of those terms by u^3 then integrate term by term

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0oh you aren't asking to find f

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0what are you asking for

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just started learning calculus and came across this problem on a practice test could u tell me specifically what this is called so i can search it up and learn it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0what is the question exactly though ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was asking for the relative minimum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just dont kno how to approach this

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0well you can find the critical numbers of f by setting f'=0 and also finding where f' dne (we don't have to worry about this case here because f' is a polynomial) f'=0 means we need to solve (x2)^2(x1)^3=0 again these x's that satisfy (x2)^2(x1)^3=0 are critical numbers

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0the critical numbers is where you can possibly have relative mins or maxes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so i just set x1=0

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so you have critical numbers x=1 or x=2 now it is possible to have a rel max or min or neither at the critical numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh kk and 1 and 2 are the critical number but the relative minimum is 1 since that is the lowest

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0I would rather not find f'' we can use f' and draw a number line in test intervals around these numbers to see if f is decreasing or increasing

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439324950653:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0well we don't know that

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0i mean yes 1 is less than 2 but that doesn't mean f(1) is a relative min

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325010596:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0I chose a random number from each of the intervals

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0that were divided by the critical numbers x=1 and x=2

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0can you find f'(0) and f'(1/2) and f'(4)?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0you actually only need the sign not the numerical value

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0for example f'(0)=(2)^4(1)^3 all we need to know is that f'(0) is negative which means f is decreasing before we get to x=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f'(0)= f'(.5)= f'(4)=+

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325170086:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0now we want to know what is happening between 1 and 2 so we need to find the sign of f'(1/2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0oh let me check didn't see you post that sorry

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325249002:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so we have no relative max

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0we only have a relative min

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0at that occurs at x=2 but the rel min value is given by f(2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0f(1) was neither a rel max or a rel min because the function did not have a switch from decreasing to increasing or the other way around

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so rel min is at x=2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats also the absolute min right?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325403944:dw because on this interval we can see the lowest value of is f(2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh lol we made a mistake

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no its okay it helped me learn thnks

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325632222:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so we have to fix somethings

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i unerstand it now its k

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439325670045:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so the graph looks something like this and no I don't know if f(2) is 0 this is just a very very rough drawing

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0so yes you are right f(1) is rel min

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0and f(2) is neither rel min or max

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0and from the graph it does appear f(1) is also a abs min

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0you were asking for examples or a site for which this is called: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx
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