Question

find maxima minima using 1st derivative test.....
y=x^3+3x^2+3x
y=(x-6)^2(9-x)

Answer 1

=3x^2+6x+3
=3(x^2+2x+1)
=(x+3)(x-1)=0
x=1 x=-3
is this correct?

Answer 2

do you know what a derivative is?

Answer 3

yes but i'm asking about maxima minima....

Answer 4

right i understand, but to get maxima and minima according to the directions you provided we need to get the derivative

Answer 5

the derivative of y=x^3+3x^2+3x
is 3x^2+6x+3 right?

Answer 6

yes it is

Answer 7

then what should i do next?

Answer 8

i'm confused in the part finding the critical points.

Answer 9

well, my shadow will tell you

Answer 10

let me help a little shessa.
take the first derivative and set it = 0
find out where it is 0 and those are the critical point
take the second derivative and set that =0

Answer 11

the guy im shadowing will tell you what to do after you take the second derivative

Answer 12

all you need is the first derivative though, since you are not looking concavity

Answer 13

ok i dont mean to confuse. Critical points= first derivative shessa

Answer 14

can you pls. answer the problems above so that i could clearly understand? pls..

Answer 15

and now that i look at this function more closely , there are no maxima or minima

Answer 16

can you pls. answer the problems above so that i could clearly understand? pls..

Answer 17

how did you know?

Answer 18

can you pls. answer the problems above so that i could clearly understand? pls..

Answer 19

pls let me understand pls.

Answer 20

x^3+3x^2+3x
d/dx(x^3+3x^2+3x)=0
3x^2+6x+3=0
x^2+2x+1=0
(x+1)^2=0
x=-1

Answer 21

this is saying first derivative= where the slope =0 because thats what we set it = to.

so at x=-1 thats just where the slope is 0

Answer 22

can you pls. answer the problems above so that i could clearly understand? pls..

Answer 23

noxipro can you explain this part pls.
x^2+2x+1=0
(x+1)^2=0
x=-1

Answer 24

i understand in getting the derivative but how did it become
x^2+2x+1=0
(x+1)^2=0
x=-1

Answer 25

x^2+(a+b)x+ab=(x+a)(x+b)
since 1=a and 1=b
a+b=2
ab=1
so my a and b are both 1

Answer 26

do you understand ?

Answer 27

am is that factoring?

Answer 28

yes it is just factoring

Answer 29

am what if its like this x^3-3x
=3x^2-3
=3x^2-3=0
what is the critical points of this?

Answer 30

3(x^2-1) is my factoring correct? then critical points are?

Answer 31

jst divide everything by 3 because 0 is on the other side
x^2-1=(x+1)(x-1)

Answer 32

ah ok tnx.am the 2nd function can you help solve it?

Answer 33

(x-6)^2(9-x)
x^2-12x+36(9-x)
9x^2-12(9)x+36(9)-x^3+12x^2-36x
d/dx=18x-12(9)-3x^2+24x-36
52x-144-3x^2=0
3x^2-42x+144=0

x^2-14x+48=0
(x-8)(x-6)=0
x=8 and x=6

Answer 34

NoxiPro explain this part x^3+12x^2-36x pls..

Answer 35

since you're foiling, you first multiply x^2 -12x +36 by 9, which is the first part of (9-x). thus:
9x^2-12(9)x+36(9)

with me so far?

Answer 36

yes

Answer 37

the second part of (9-x) is the -x. and you MUST carry the negative with it.
you then multiply the original x^2-12x+36 by that -x.
this gives you:
x^2(-x)-12x(-x)+36(-x)
which simplifies to:
-x^3+12x^2-36x

Answer 38

ah ok i got it now thank you very much