if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has

- anonymous

if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has

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- 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

so multiply out (u-1)^2 then multiply each one of those terms by u^3
then integrate term by term

- myininaya

oh you aren't asking to find f

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- myininaya

what are you asking for

- 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

what is the question exactly though ?

- anonymous

i was asking for the relative minimum

- anonymous

i just dont kno how to approach this

- 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

the critical numbers is where you can possibly have relative mins or maxes

- anonymous

oh so i just set x-1=0

- myininaya

and x-2=0

- 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

oh kk and 1 and 2 are the critical number but the relative minimum is 1 since that is the lowest

- 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

|dw:1439324950653:dw|

- myininaya

well we don't know that

- myininaya

i mean yes 1 is less than 2
but that doesn't mean f(1) is a relative min

- myininaya

|dw:1439325010596:dw|

- myininaya

I chose a random number from each of the intervals

- myininaya

that were divided by the critical numbers x=1 and x=2

- myininaya

can you find f'(0)
and f'(1/2)
and f'(4)?

- myininaya

you actually only need the sign not the numerical value

- 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

f'(0)=-
f'(.5)=-
f'(4)=+

- myininaya

|dw:1439325170086:dw|

- 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

oh let me check
didn't see you post that sorry

- myininaya

|dw:1439325249002:dw|

- myininaya

so we have no relative max

- myininaya

we only have a relative min

- myininaya

at that occurs at x=2
but the rel min value is given by f(2)

- 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

so rel min is at x=2?

- myininaya

yep

- anonymous

thats also the absolute min
right?

- myininaya

|dw:1439325403944:dw|
because on this interval we can see the lowest value of is f(2)

- anonymous

oh lol we made a mistake

- myininaya

what is the mistake

- anonymous

its 1.5

- myininaya

lol yes

- anonymous

so 1 is minimum

- myininaya

that was dumb sorry

- anonymous

no its okay it helped me learn thnks

- myininaya

|dw:1439325632222:dw|

- myininaya

so we have to fix somethings

- anonymous

i unerstand it now its k

- myininaya

|dw:1439325670045:dw|

- 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

so yes you are right f(1) is rel min

- myininaya

and f(2) is neither rel min or max

- myininaya

and from the graph it does appear f(1) is also a abs min

- myininaya

you were asking for examples or a site for which this is called:
http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

- anonymous

appreciate it

- myininaya

np

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