Question

Determine where the function is concave upward and downward, and list all the - value(s) of the point(s) of inflection:
f(x)=4x^3+2x^2+2x-9

Answer 1

where are you stuck on this one?

Answer 2

i got (-inf,-1/6)
(-1/6,inf)
whats the inflections point

Answer 3

for the inflection point you plug your critical point from the second derivative back into the second derivative equation - since the inflection point is where the concavity changes.

Answer 4

thus f'(x) = 12x^2+4x+2 right?

Answer 5

yes

Answer 6

so f'' is just 24x+4 ?

Answer 7

correct

Answer 8

set the second derivative to zero to get the critical point ?

Answer 9

-1/6

Answer 10

plug that back into f(x) sorry i said second deriv earlier. but if you want coordinates.. you just have to plug back into f(x)

Answer 11

i did but for some reason my answer is coming up wrong

Answer 12

sec working it on paper myself. sorry.

Answer 13

nevermind i got it!
it was only asking for the x values i was putting in the point

Answer 14

do you mind helping me with another problem

Answer 15

ooh lol.

Answer 16

not at all.. go ahead.

Answer 17

Suppose that the position of a particle as a function of time (in seconds)is given by the formula s(t)=5.5+5t^3-t^5 t>0

find The time at which the velocity is maximum

Answer 18

ok so same thing here. the the derivative of s(t) to get velocity right?

Answer 19

yes

Answer 20

then find the critical points of the derivitive

Answer 21

i got x= -sqrt(3) x=0 and x=sqrt(3)

Answer 22

yes that is the velocity. since the domain is x>0 the negative number is not what we want.

Answer 23

i already fouund velocity and acceleration.. im having trouble finding when velocity is maximum

Answer 24

when you find the critical point for the first derivatives.. its either a local max, local min, abs max, or abs min.

Answer 25

so the max is going to occur when v(t)=0 or t=sqrt(3) as you said.

Answer 26

v(0)=5.5 and v(sqrt(3) is 15.8923

Answer 27

you only need to set the velocity v(t) = 0. which you did at got 1.73.... or sqrt3. so @ time 1.73 seconds, the particle is moving at its maximum velocity. if you need the position then you plug sqrt3 back into s(t).

Answer 28

its coming up wrong

Answer 29

you using MML?

Answer 30

?

Answer 31

sorry if im not much help.. im also a calc student this semester. Do you think if we set the acceleration = to 0 we can see what time the velocity is changing? thus my thinking is that when a(t)=0, the velocity is at a max? maybe it isn't right. i got t=1.22 seconds. I suppose it would help to graph it. I have not done that.

Answer 32

its alright, ill just skip that part for now

Answer 33

@Luis_Rivera might be able to help.. he is good.

Answer 34

thanks