Question

For the curve y=2-12x+9x^2-2x^3:

i) Find the intervals of increase and decrease.
ii) Find the local maximum and minimum values.
iii) Find the intervals of concavity.
iv) Find any points of inflection.
v) Sketch the curve.

Answer 1

The first step would just be finding the derivative, since most of the questions can be answered with information about it.

Answer 2

y'=-6x^2+18x+12

Answer 3

y' = -12 + 18x - 6x^2

Answer 4

**-12

Answer 5

solve y' = 0 to find max/minm

Answer 6

how to do you solve y'=0?

Answer 7

Set "-12 + 18x - 6x^2 = 0"
solve for x

Answer 8

ok

Answer 9

I really need to understand how to do part i) if you guys can help me out with that one first

Answer 10

well, finding the intervals also involves finding the 0's of the derivative. We'll use the points where the derivative is 0 as reference points and test x-values around them to find if the function is increasing or decreasing at those points (since when y' = 0, the graph could either change signs or stay the same, so these points are important to look at)

Answer 11

ok to do y'=o what exactly do I do

Answer 12

We'll write out what we have for y'

-6x^2 + 18x - 12 = 0 (written in proper general form)
we can factor out a monomial -6 and divide it off since its just a constant
x^2 - 3x + 2 = 0
then we factor the new expression since -1 * -2 = 2 and -1 + -2 = -3
(x - 1)(x - 2) = 0 ab = 0; a = 0 or b = 0
x - 1 = 0 and x - 2= 0
x = 1, x = 2

Answer 13

Ok I get that

Answer 14

so it is (x-1)(x+2)? right

Answer 15

good Question. :)

Answer 16

it factors to (x-1)(x-2) = 0
we solve for the x-values that will get us 0
x - 1 = 0
x = 1

x - 2 = 0
x = 2

These values will be important to mark out our intervals.



We'd test x-values around those intervals in the derivative to determine if it is postivie or negative for that interval.
ie. we could use x=0, x=1/2, and x=3

Answer 17

* that would tell us if that interval is increasing or decreasing, since a positive derivative indicates increase and a negative derivative indicates decrease.

Answer 18

ok so where do we substitute these test values and how do we get the test values?

Answer 19

We substitute them into the derivative y' = -6x^2 + 18x - 12 and find the y' value

They're pretty much our own choice. We can choose any values as long as they're not x=1 or x=2 (we already know the derivative is 0 there, no need to test them.)
If you wanted, we could use x=26 rather than x=3 but it wouldn't be as nice to calculate. :)

Answer 20

when do we know when to stop testing?

Answer 21

we only need to test one value of x in each interval
x < 1, 1 < x < 2, and 2 < x

Answer 22

oo so because its a parabola the intervals would be positive for 1<x<2 and negative for x<1 and x>2 right?

Answer 23

My bad its -2

Answer 24

yeah, that would be correct
graphically, you can tell pretty easily
my method is just the analytical way, if you didn't have he graph.
it is actually positive 2, not -2.
since (x-1)(x-2) given x=2 => (2-1)(2-2) = 1*0 = 0 works

Answer 25

i thought it was (x-1)(x+2)?

Answer 26

oo ye nvm ur right

Answer 27

kk
so, we have the intervals of change.
x<1 decreasing
1<x<2 increasing
x>2 decreasing

To identify the local maxima and minima, we find them by using those points we found for the intervals (x=1 and x=2) and look at the intervals just around the points
if the function goes from decreasing to increasing, we have a minimum value at the point
if the function goes from increasing to decreasing, we have a maximum value at the poin

Answer 28

but since its a parabola wont it only have 1 min/max?