Question

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

Answer 1

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

Answer 2

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

Answer 3

oh you aren't asking to find f

Answer 4

what are you asking for

Answer 5

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

Answer 6

what is the question exactly though ?

Answer 7

i was asking for the relative minimum

Answer 8

i just dont kno how to approach this

Answer 9

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

Answer 10

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

Answer 11

oh so i just set x-1=0

Answer 12

and x-2=0

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

well we don't know that

Answer 17

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