Question

If
f'(x) = x^3 (x-2)^4 (x-3)^2 ,
then f(x) has a relative maximum at what value(s) of x ?

Answer 1

@SolomonZelman can you please help

Answer 2

or anyone

Answer 3

When f'(x) =0?

Answer 4

x= 0 ?

Answer 5

2more

Answer 6

im not sure

Answer 7

Confirm: \(f'(x) = x^3(x-2)^4(x-3)^2\) right? I mean f' not f, right?

Answer 8

a bit more to it than that
at x = 0 the derivative goes from being negative to positive, so the function goes from decreasing to increasing
that give you a relative minimum there, not a maximum

Answer 9

that is what they gave me

Answer 10

I am really confused

Answer 11

your derivative has 3 zeros
it is clear what they are?

Answer 12

no I am not sure , how do I know that

Answer 13

wait yes , one for x^3 , another for (x-2)^4 and another for (x-3)^2 ?

Answer 14

yes of course, it is in factored form, the zeros are displayed for you

Answer 15

0 -2 and -3 ?

Answer 16

no

Answer 17

close

Answer 18

yeah I dont know

Answer 19

\[x-2=0\\
x=?\]

Answer 20

0 2 and 3

Answer 21

whew

Answer 22

now those are your candidates for local max and min

Answer 23

ok

Answer 24

i am pretty sure there is no local max but there certainly is a local min

Answer 25

and can you show me how to do that ? I am not sure what to do

Answer 26

you have a couple choices
if the derivative changes sign from negative to postive, that means the function goes from decreasing to increasing
that means you have a local min

Answer 27

yeah correct and if it does the opposite then we have a max

Answer 28

right
now
it should be clear that \((x-2)^4\) and \((x-3)^2\) are always positive, (unless they are zero) because the exponents on each term are even

Answer 29

that means the derivative does not change sign at 2 and at 3, it just kisses the x axis and then continues on its way
that in turn means the function does not go from increasing to decreasing or vice versa, just flattens out there

Answer 30

the only change in sign comes from the \(x^3\) term
that goes from negative to positive as x goes from negative to positive, i.e at \(x=0\)

Answer 31

because it is odd

Answer 32

exactly

Answer 33

in plainer english your derivative is negative if x is negative and positive if x is positive, except when it is zero

Answer 34

which might be confusing because you are asked for a local max, and there is not one

Answer 35

because it asks for a relative max for f(x) not f'(x) because the equation they gave us is f prime but it is asking for the relative max of the actual f(x) do you get it ?

Answer 36

yes we are analyzing the behaviour of the function from the derivativef

Answer 37

since the derivative is negative for \(x<0\) and positive for \(x>0\) the function has a local MINIMUM at \(x=0\) and no local max

Answer 38

oh , so thats is what I say?

Answer 39

that is what i say

Answer 40

you can say it too if you like