Question

Find the intervals of concavity and the inflection points of f(x)= x sqrt(x+3). I found my first derivative but I don't know if I'm doing the second derivative correctly. Help please!

Answer 1

Sure, so first off, do you understand why taking a derivative of anything would tell you where to find concavity and inflection points?

Answer 2

Just explain as best as you can what you think you're doing and I'll help you out! =D

Answer 3

Please SHOW your 1st Derivative.

Answer 4

Here's my first derivative \[\frac{ 3(x+2)}{ 2\sqrt{x+3}}\]

Answer 5

okay when I solve to get my second derivative I get this. I don't know if it's right so far. \[\frac{ 12x+36 }{ \sqrt{x+3}} - \frac{ 6x+2 }{ \sqrt{x+3}} = \frac{ 6x+38 }{\sqrt{x+3} }\]

Answer 6

Not really sure how you managed that 2nd Derivative. Can you describe your methodology? Quotient Rule? What?

Answer 7

I used the quotient rule. Oh but I just realized I forgot the \[g^2\]from the quotient rule so my answer must be wrong

Answer 8

Should be: \(\dfrac{3}{2}\dfrac{\sqrt{x+3}\cdot\dfrac{d}{dx}(x+2) - (x+2)\cdot\dfrac{d}{dx}\sqrt{x+3}}{x+3}\).

Please note how removing the constants up front made life a whole lot easier.

Answer 9

Thats makes things a lot clearer! But how did you remove the radical from \[\sqrt{x+3}\] in the bottom.

Answer 10

\(g^{2}\), remember?

Answer 11

Oh right! okay, let me solve it now to see if im doing it right.

Answer 12

\[\frac{ 3 }{ 2}\frac{ \sqrt{x+3}-(x+2)(\frac{ 1 }{ 2 }) (x+3)^{-\frac{ 1 }{ 2 }}}{{x+3} }\] Okay so i know in order to make \[(x+3)^{\frac{ -1 }{ 2}}\] I would have to bring it to the bottom would that be \[x+3\sqrt{x+3}\] ?

Answer 13

I don't know what to do after this.

Answer 14

Good try. You can work on your bookkeeping, later. You should get:

\(\dfrac{3}{4}\cdot\dfrac{x+4}{(x+3)^{3/2}}\) for the 2nd Derivative.

Answer 15

Hmm, okay so from there how would I find the inflection point?

Answer 16

What makes the 1st Derivative 0 or undefined?
What makes the 2nd Derivative 0 or undefined?

Answer 17

for the first derivative I got x= -2 is that correct?

Answer 18

and the second derivative x=-4?

Answer 19

We should point out that the Domain of f(x) is \(x \ge -3\). Do you see why this is so?

How does this make you feel about x = -4 for a zero of the 2nd derivative?

Answer 20

Why is the domain of\[f(x) \ge -3\] ?

Answer 21

\(\sqrt{x+3}\) Does x = -5 work in there?

Answer 22

No it would make it undefined.

Answer 23

So, what is the Domain? What values work in that square root?

Answer 24

-2 would work?

Answer 25

or does it have to equal to zero?

Answer 26

It's continuous. \(x \ge -3\)

Answer 27

OH! okay I understand.

Answer 28

so then -4 would not work for the bottom. but when i set the top to 0 it gives me x= -4

Answer 29

Yup. It's not in the Domain. Just ignore it.

Next important question: What is the Domain of \(f'(x)\)?

Answer 30

Would it be the same thing ? \[x \ge -3\]

Answer 31

Put x = -3 in there and see how that goes.

Answer 32

It gives me 0

Answer 33

For the DENOMINATOR! Is that okay?

Answer 34

hmm so then is it just \[x >-3\] ?

Answer 35

That is correct. Same thing for the 2nd Derivative.

Recap:

f(x) Domain \(x \ge -3\)

f'(x)

Zero: x = -2
Fails to exist: x = -3

f"(x)

Zero: None
Fails to exist: x = -3

We are ready to identify the Critical Points. What are they?

Answer 36

x = -2 ?

Answer 37

And x = -3. Don't miss these oddballs.

Answer 38

Oh okay. Got it.

Answer 39

Okay, one last thing. Classify the Critical Points.

Answer 40

Do I plug them into my second derivative?

Answer 41

Or my original function?

Answer 42

No, that is not likely to work, since x = -3 won't go!

Think about them.

x = -2
f'(-2) = 0 -- This this could be a min or a max.
f"(-2) > 0
This is a minimum value. It turns out to be a Global minimum. The second derivative is everywhere positive, so this graph won't be doing any more tricks.

x = -3
What do you think?

Answer 43

So x= -3 can't be a minimum or a maximum?

Answer 44

Not so. The derivative doesn't tell you anything, but it still can be. You must check places like this by yourself.

What is the sign of the derivative just shy of x = -3?
We already established that the 2nd derivative is positive everywhere it exists.

Answer 45

We often encounter this kind of problem with artificial limits on the Domain. The problem statement suggests a function, but then allows x only between two values, say 1 and 2. The derivative knows nothing of the artificial limitations. You must examine these oddities by yourself.

This problem has a more natural endpoint, but the havoc with the derivatives is the same.

Answer 46

f(-3) = 0
f'(-3+a hair) < 0
f"(-3+a hair) > 0

This guy is a local maximum.

Answer 47

Now I understand. So would it be concave up on \[(-3, \infty)\] ?

Answer 48

Yes.

Answer 49

Wow you helped me so much with this. I actually understand it now. I just have to work on finding the second derivative. Thank you so much for your help and time. I really appreciate it! :)

Answer 50

No worries. It's what we strive to do around here. Good working hanging in there!