anonymous
  • anonymous
if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has
Mathematics
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anonymous
  • anonymous
if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has
Mathematics
schrodinger
  • schrodinger
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myininaya
  • myininaya
so you need help integrating then \[\int\limits (x-2)^2(x-1)^3 dx \] Let u=x-1 then du=dx if u=x-1 subtracting one on both sides gives: u-1=x-2 so you have \[\int\limits_{}^{}(u-1)^2u^3 du\] this is just to make the multiplication less harsh we still have to multiply
myininaya
  • myininaya
so multiply out (u-1)^2 then multiply each one of those terms by u^3 then integrate term by term
myininaya
  • myininaya
oh you aren't asking to find f

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myininaya
  • myininaya
what are you asking for
anonymous
  • anonymous
i 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
  • myininaya
what is the question exactly though ?
anonymous
  • anonymous
i was asking for the relative minimum
anonymous
  • anonymous
i just dont kno how to approach this
myininaya
  • myininaya
well 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 (x-2)^2(x-1)^3=0 again these x's that satisfy (x-2)^2(x-1)^3=0 are critical numbers
myininaya
  • myininaya
the critical numbers is where you can possibly have relative mins or maxes
anonymous
  • anonymous
oh so i just set x-1=0
myininaya
  • myininaya
and x-2=0
myininaya
  • myininaya
so 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
  • anonymous
oh kk and 1 and 2 are the critical number but the relative minimum is 1 since that is the lowest
myininaya
  • myininaya
I 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
  • myininaya
|dw:1439324950653:dw|
myininaya
  • myininaya
well we don't know that
myininaya
  • myininaya
i mean yes 1 is less than 2 but that doesn't mean f(1) is a relative min
myininaya
  • myininaya
|dw:1439325010596:dw|
myininaya
  • myininaya
I chose a random number from each of the intervals
myininaya
  • myininaya
that were divided by the critical numbers x=1 and x=2
myininaya
  • myininaya
can you find f'(0) and f'(1/2) and f'(4)?
myininaya
  • myininaya
you actually only need the sign not the numerical value
myininaya
  • myininaya
for 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
  • anonymous
f'(0)=- f'(.5)=- f'(4)=+
myininaya
  • myininaya
|dw:1439325170086:dw|
myininaya
  • myininaya
now we want to know what is happening between 1 and 2 so we need to find the sign of f'(1/2)
myininaya
  • myininaya
oh let me check didn't see you post that sorry
myininaya
  • myininaya
|dw:1439325249002:dw|
myininaya
  • myininaya
so we have no relative max
myininaya
  • myininaya
we only have a relative min
myininaya
  • myininaya
at that occurs at x=2 but the rel min value is given by f(2)
myininaya
  • myininaya
f(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
  • anonymous
so rel min is at x=2?
myininaya
  • myininaya
yep
anonymous
  • anonymous
thats also the absolute min right?
myininaya
  • myininaya
|dw:1439325403944:dw| because on this interval we can see the lowest value of is f(2)
anonymous
  • anonymous
oh lol we made a mistake
myininaya
  • myininaya
what is the mistake
anonymous
  • anonymous
its 1.5
myininaya
  • myininaya
lol yes
anonymous
  • anonymous
so 1 is minimum
myininaya
  • myininaya
that was dumb sorry
anonymous
  • anonymous
no its okay it helped me learn thnks
myininaya
  • myininaya
|dw:1439325632222:dw|
myininaya
  • myininaya
so we have to fix somethings
anonymous
  • anonymous
i unerstand it now its k
myininaya
  • myininaya
|dw:1439325670045:dw|
myininaya
  • myininaya
so 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
  • myininaya
so yes you are right f(1) is rel min
myininaya
  • myininaya
and f(2) is neither rel min or max
myininaya
  • myininaya
and from the graph it does appear f(1) is also a abs min
myininaya
  • myininaya
you were asking for examples or a site for which this is called: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx
anonymous
  • anonymous
appreciate it
myininaya
  • myininaya
np

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