Question

CRITICAL POINTS

This is the problem
x^2 / (x^2+3)

I found the first derivative 6x/(x^2+3)^2
But I don't know what to do next to find the zeroes. I know I have to equal the derivative to 0 and I did but then I got lost. I'd appreciate any help, thank you

Answer 1

On the closed interval [-1,1]

Answer 2

Can you help me out?

Answer 3

Derivative looks good :)\[\large\rm 0=\frac{6x}{(x^2+3)^2}\]Multiplying through by our denominator gives us,\[\large\rm 0=6x\]

Answer 4

How did you get 6x? Sorry I need to refresh my algebra

Answer 5

Multiplying both sides by \(\large\rm (x^2+3)^2\)

Maybe a good shortcut to remember is just to `set your numerator equal to zero`.

Answer 6

Now, something you should realize though...
We also must set our denominator equal to zero.

\(\large\rm 0=(x^2+3)^2\)
This corresponds to a `different type of critical point`,
but it's still a critical point.

It doesn't refer to locations of horizontal tangency,
but instead locations of vertical tangency.

Answer 7

Ohhhh since the zero is on the other side it cancels out. Yeah I need to remember setting it equal to zero, Thanks!! So the critical point would be x=-6

Answer 8

No no :O woops!
0=6x, the 6 is `multiplying` x.
So you have to `divide`.

Answer 9

Right. So x=0 again I need to remember setting it equal to zero

Answer 10

Now, if I wanted to find the absolute extrema I need to plug in 0 into the derivative or the original equation?

Answer 11

Had to leave class early so I kind of had to learn it alone. It sucks

Answer 12

Critical points refer to locations of horizontal tangency.
`The slope of the function at these critical points is zero`.

So if you plugged your critical point back into the derivative, it would give you zero. That's what we figured out by setting the derivative equal to zero.

So that won't do us any good :)
Plug it back into the original function get an exact location (coordinate pair) of your extrema.

Answer 13

If I plug 0 into the original equation I get (0,0)

Answer 14

So this min or max point (whatever it may be) goes through the origin?
Yup looks correct! :)

Answer 15

Don't I have to plug in the intervals too? It's set on a closed interval [-1,1] don't I have to plug them too?

Answer 16

Trying to find the maxima

Answer 17

Oh oh oh oh.. yes sorry >.<

When you're given a closed interval,
you check for critical points,
then you take all of the important points that you have:
critical points and end points, and plug them into the original function,
and compare their outputs.

Answer 18

Mmm but I get (-1,1/4) and (1,1/4) which one will be the maxima? I mean they're both the same

Answer 19

Maximum: (-1, 1/4) and (1, 1/4)
Minimum: (0,0)

Yes :) good job!

We list critical points by their x-coordinate,
but extrema as a coordinate pair.

Answer 20

Oh I see. Thanks so much! I have another problem which is just finding a derivative that I wouldn't find if you have time?

Answer 21

Couldn't*

Answer 22

It's f(x)= x(sqr 4-x)

Answer 23

I'm on a phone so I can't use the symbols

Answer 24

\[\large\rm f(x)=x\cdot\sqrt{4-x}\]Hmm any ideas how to start this one?
We have two things multiplying together, anything clicking? XD

Answer 25

Lol yeah I mean I got (1)(sqr 4-x) +

Answer 26

And then that part I didn't get lol I know it's product rule and.... chain?

Answer 27

I think it's called a composite derivative or something like that

Answer 28

Early on, it might be a good idea to spend an extra step "setting up" your product rule. If you're comfortable skipping that, it's fine.\[\large\rm f'(x)=(x)'\sqrt{4-x}+x(\sqrt{4-x})'\]

Answer 29

\[\large\rm f'(x)=(1)\sqrt{4-x}+x(\sqrt{4-x})'\]

Answer 30

Yeah Ima confortable skipping it but I got all messed up trying to find the derivative of the square inside the product

Answer 31

Im*

Answer 32

You can convert your root to a 1/2 power if you like,
but honestly, the square root comes up soooo oftne,
it's really worth your time to memorize the square root derivative.

Answer 33

I see

Answer 34

\[\large\rm \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt x}\]Derivative of square root is one over two square roots.

Answer 35

So yes, you'll have to apply the chain rule as extra step since our inner function is `more than just x`,\[\large\rm \frac{d}{dx}\sqrt{4-x}\quad=\quad \frac{1}{2\sqrt{4-x}}\cdot (4-x)'\]

Answer 36

So how would you set it up? The chain rule and the product rule together for this problem

Answer 37

The chain part of the problem would be: (x) • 1/2(sqr 4-x) • (-1) ...... (-1) coming from the derivative of 4-x

Answer 38

Well, you chain every time you take a derivative.
That let's us know that product rule will be applied first, ya?
Because the "setting up" of our product rule isn't actually the process of taking a derivative.

Yes that looks good.
Lemme recap just in case it will help...

Answer 39

Alright. I wish I could send pictures through here

Answer 40

Oh you can't see the LaTex formatted? :(
Ugh, I was gonna do it in color >.< heh

Answer 41

What do you mean? I pretty much can see everything I just can't send it

Answer 42

Haha I misread your last line, read it as "I wish I could `see` pictures through here" XD lol

Answer 43

Lol I figured

Answer 44

So I like to use this color theme:
Blue are derivatives that need to be taken,
Orange is when we finish them.

So we set up our product rule,
\[\large\rm f'(x)=\color{royalblue}{(x)'}\sqrt{4-x}+x\color{royalblue}{(\sqrt{4-x})'}\]Take our first derivative,\[\large\rm f'(x)=\color{orangered}{(1)}\sqrt{4-x}+x\color{royalblue}{(\sqrt{4-x})'}\]Take the other one, creating a chain,\[\large\rm f'(x)=\color{orangered}{(1)}\sqrt{4-x}+x\color{orangered}{\frac{1}{2\sqrt{4-x}}}\color{royalblue}{(4-x)'}\]Finishing up,\[\large\rm f'(x)=\color{orangered}{(1)}\sqrt{4-x}+x\color{orangered}{\frac{1}{2\sqrt{4-x}}}\color{orangered}{(-1)}\]

Answer 45

And then maybe getting a common denominator from there.
But whatever.

Answer 46

So one tiny thing to notice is that the chain is only multiplying the second term, so we don't end up distributing that -1 to each term or anything like that.

Answer 47

Yeah I noticed. Now how would the next step look after multiplying the 1, -1, and x? Cause I did it on my own but i got one whole fraction

Answer 48

(Sqr 4-x) - x all over (sqr 4-x)

Answer 49

\[\large\rm f'(x)=\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\]Hmm I guess that first term needs a 2sqrt(4-x) to have the same denominator, ya?\[\large\rm f'(x)=\left(\frac{2\sqrt{4-x}}{2\sqrt{4-x}}\right)\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\]Giving us,\[\large\rm f'(x)=\frac{2(4-x)}{2\sqrt{4-x}}-\frac{x}{2\sqrt{4-x}}\]combining the fractions,\[\large\rm f'(x)=\frac{2(4-x)-x}{2\sqrt{4-x}}\]Looks like the square root disappears from the numerator, ya?

Answer 50

Mmmm yeah it disappears but we still have it in the denominator

Answer 51

So ya I guess you could go a step further,
distribute the 2, and then combine like-terms.\[\large\rm f'(x)=\frac{8-3x}{2\sqrt{4-x}}\]

Answer 52

Oh what I did was I put everything under the same numerator but It was wrong

Answer 53

That would be the final form of the derivative?

Answer 54

Oh I see now

Answer 55

CRitical point would be 8/3? Lol

Answer 56

Ya, that's a good looking final answer.
The earlier one was fine also though,\[\large\rm f'(x)=\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}\]Just depends how far your teacher wants you to simplify things.

Answer 57

Yeah he wants us to go all the way when it comes to simplifying

Answer 58

You might have missed the subtle point I made earlier,
Yes, 8/3 would end up being a critical value for the function.
But we also have to set the denominator equal to zero to find other critical values.\[\large\rm 0=2\sqrt{4-x}\]

Answer 59

The reason we didn't worry about this in the last problem is because: The value we would have ended up with was `not in the domain of our original function`. The function and derivative had the same factor in the denominator, so the solution wasn't allowed.

That's not the case for this problem though,
you'll need to solve this equation to find the other critical point.

Answer 60

Yeah but I thought if we multiplied both sides by the denominator, it'll end up canceling out cause the 0 is on the other side. Then we end up having 8-3x=0

Answer 61

It's a different type of critical point :)
It's a weird one.

Answer 62

Well the problem added that x<3, they didn't give me an interval just asking for critical points

Answer 63

Idk if the numerator fits x<3

Answer 64

Denominator***

Answer 65

So you found this critical point.
The slope of the line tangent to the curve is zero (horizontal).