Question

Use derrivatives to determine the intervals on which f(x)=x^3+3x^2-9x+3 is increasing and the intervals where it is decreasing.

Answer 1

you would need to find the stationary points...

which needs 1st derivative then solving the 1st derivative.

find the second derivative

and test the stationary points

solve the 2nd derivative for x..

check to see if there is a change in concavity either side of your solution... if so the solution is a point of inflection.

Answer 2

The intervals on which dy/dx >0 , f(x) will be increasing and
The intervals on which dy/dx <0 , f(x) will be decreasing .
what you got for dy/dx ?

Answer 3

So far After taking the first derrivative
and setting it equal to zero I got 1/2 and -3/2..but that doesn't really look right.

Answer 4

setting first derivative =0 will give you extrema, which you don't exactly want.
3x^2+6x-9 >0 (or x^2 +2x-3>0) will give you intervals on which f(x) is increasing.
similarly,
3x^2+6x-9 <0 (or x^2 +2x-3<0) will give you intervals on which f(x) is decreasing.

Answer 5

Okay so for the increasing one can I just select a number greater than 0 such as 9 and add three then set x^2 equal to 12 and then 2x=9?

Answer 6

how come x^2=12 ??
x^2 +2x-3>0
(x+3)(x-1) >0
can you find intervals from this ?

Answer 7

no ?

Answer 8

Alright so
I get x=-3 and x=1

Answer 9

thats for (x+3)(x-1) = 0
and not (x+3)(x-1) >0 , you know how to solve this inequality ?

Answer 10

Ohh, no I do not.

Answer 11

for (x+3)(x-1) to be positive (>0)
either both (x+3) and (x-1) must be positive
or both (x+3)and (x-1) must be negative, right ?

Answer 12

right.

Answer 13

so, x+3 >0 and x-1 >0 OR x+3<0 and x-1 <0

x>-3 and x>1 OR x<-3 and x<1
got this ?

Answer 14

Yess, I do :)

Answer 15

can you simplify this : x>-3 and x>1 ?

Answer 16

I don't know..

Answer 17

okk..
x>-3 means all values of x greater than -3
x>1 means all values of x greater than 1
' and ' means common elements
so, all elements for x>1 are common only.
so, x>-3 and x>1 -----> x>1 only
got this ?

Answer 18

Okaayy

Answer 19

what about x<-3 and x<1 ?

Answer 20

x<-3?

Answer 21

thats correct :)
so, now you have
x>1 or x<-3
thats your interval for which f(x) is increasing.
in interval notation, it'll be (-infinity, -3)U(1,infinity)
do similar process for decreasing interval.

Answer 22

Ahh...Okayy! Thank you very very much!

Answer 23

welcome ^_^
you can ask me if you get stuck ....or want to verify your answer.