Question

Find the derivative of
(1-x)^2 (1+x)^3

Then find relative min or max and increasing and decreasing

Answer 1

\[\Large \color{red}{\text{Welcome To OpenStudy :)}}\]

So it looks like we'll need to start by applying the product rule:\[\large \left[(1-x)^2 (1+x)^3\right]'=\color{royalblue}{\left[(1-x)^2\right]'}(1+x)^3+(1-x)^2\color{royalblue}{\left[(1+x)^3\right]'}\]Understand the setup? We'll need to take the derivative of the blue portions.

Answer 2

Yes I understand the setup. Would you use the chain rule to find the derivatives?

Answer 3

So would it be 2(1-x) times -1 ?

Answer 4

For the first term? Yes, very good.

Answer 5

\[\Large \color{orangered}{-2(1-x)}(1+x)^3+(1-x)^2\color{royalblue}{\left[(1+x)^3\right]'}\]

Answer 6

Then for the second term it would just be 3(1+x)?

Answer 7

\[\Large \color{orangered}{-2(1-x)}(1+x)^3+(1-x)^2\color{orangered}{3(1+x)^2}\]Woops, careful with the power rule there :)

Answer 8

OH oops I knew that. Okay okay so now I can take out an (1+x) and a (1-x) ??

Answer 9

Hmm looks like you can take out (1+x)^2 from each term.

Answer 10

along with the (1-x) i mean

Answer 11

Right so I am left with (1-x)(3) + (1+x)(-2)

Answer 12

But with the (1+x)^2 and the (1-x) still there just on the outside

Answer 13

\[\Large (1-x)(1+x)^2\left[-2(1+x)+3(1-x)\right]\]Ah yes, that sounds right. Good factoring.

Answer 14

Then in the brackets should I factor the -2 and 3 in? I forgot how to get the final answer

Answer 15

Yah let's simplify the stuff in the square brackets, shouldn't be too much work.
Distribute the -2 and 3.

Answer 16

Okayyyyy so 1-5x

Answer 17

\[\Large f(x)=(1-x)^2(1+x)^3\]\[\Large f'(x)=(1-x)(1+x)^2(1-5x)\]

Ok looks good so far :)

Answer 18

OHHHHH okay so then I just equal each (1+x), (1-x) (1-5x) to zero?

Answer 19

To find critical points? Yes, good.
We set the first derivative equal to zero.\[\Large 0=(1-x)(1+x)^2(1-5x)\]The Zero Factor Property tells us that we can set each individual factor equal to zero to solve for x in each case.

Answer 20

Yeaah so x= 1, -1, and 1/5

Answer 21

Ok your critical points look good.

What method are you familiar with for finding max/min, increasing decreasing?
Do you usually draw out a chart with the + and - stuff?
I had learned to do it with drawing out a number line, so I'm just curious :3 heh

Answer 22

I usually do it with a chart but I was taught both ways with a number line & chart. Chart is easier... in my opinion

Answer 23

I plug the points back into the derivative right?

Answer 24

Umm no.
I mean yes, eventually we'll need to do that.

Because we want to list our max/min as ordered pairs.
So if you want to get those 3 ordered pairs out of the way now, that's fine.

I usually figure out which values are my max/min before worrying about the y values.

Answer 25

Oh back into your derivative? Doing that will give us 0.
We need to plug in `test points`, points on the left and right side of our critical points to find out what is happening around that point.

Answer 26

Yeah no.. that's what I meant, the test points lol

Answer 27

Hah :3 yah cool.

So like for the left most side, we could try,\[\Large f'(-2)\]We only care about the `sign` of our answer that it spits out. So don't worry about doing too many calculations. Just try to keep track of the sign.

Answer 28

For example, \(\Large (1+x)^2\) is always positive, so we can completely ignore that term for this process.

Answer 29

Okay so I got positive, positive, negative, positive when I tested out -2, 0, .5, and 2

Answer 30

Ok great!

Answer 31

I like drawing out the number line myself :)
Makes it easier for me to identify max/min.
I just look for where the arrows meet.

Answer 32

Yup max at 1/5 and min at 1

Answer 33

Then you plug those point back into the original equation

Answer 34

Ah yes, to get the ordered pairs.

Answer 35

I am unsure if they are right but max of 1.1 occurs at x=1/5
and min of 0 occurs at x=1

Answer 36

Yah that sounds good! :) Here is a graph you can use to verify.
https://www.desmos.com/calculator/mvx1mvqh7q
You can hover over the critical points to see the ordered pairs.

Answer 37

Oh cool! I understand it now thank you so much!!
Just for future reference when I am given a problem like this, I use the product rul then the chain rule to find the derivative then I factor out something. Then I find the critical points by equalling them to zero then find the inc/dec min/max with test points

Answer 38

The process will always follow that pattern, yes:
~Take derivative
~Set derivative equal to zero to find critical points
~Plug test points into derivative function to determine max/min, increasing/decreasing
~Plug max/min into original function to get y values for them

Some things might be a little different along the way.
Like you might not always start by using the product rule and then chain rule.
It really depends what your function looks like.

And then sometimes things can get a tad tricky when they throw trig functions at you since you need to remember your special angles.