NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3)
* This is a calc problem

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- freckles

Have you considered finding the derivative?

- UsukiDoll

we need the first derivative test... wait we need to find critical points first!

- freckles

f'>0 tells us f is increasing
f'<0 tells us f is decreasing
Find the critical numbers and consider the domain.

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## More answers

- UsukiDoll

what @freckles said is correct... use product rule

- anonymous

I already did all that, and I got (sqrt x+3) + x/(2(sqrt x +3)). But I just dont know how to go about solving for the critical values

- freckles

I don't think we need to find the x-intercepts to find where the function is increasing/decreasing.

- anonymous

is it just x= -3?

- UsukiDoll

noooooo....................... freckles we need the critical points first or we can't go further T_T

- UsukiDoll

sorry mary that comment was at freckles

- freckles

\[f'(x)=\sqrt{x+3}+\frac{x}{2 \sqrt{x+3}} \\ \text{ solve } f'(x)=0 \text{ \to find the critical numbers }\]

- Zale101

To get the critical points, we first need to set the derivative to zero

- anonymous

Yea I did that already guys!!!

- Zale101

and solve for x. Those x's are the crtical points.

- freckles

good then what did you for the x's?

- anonymous

I just need help in solving for x

- freckles

\[\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}\]
find common denominator

- anonymous

No, we find the first derivative and then find the critical points

- freckles

\[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0\]

- anonymous

hey freckles, how did you get +x in the numerator?

- freckles

(sqrt x+3) + x/(2(sqrt x +3)
isn't this what you got for f'?

- anonymous

OH, K i see it

- freckles

\[f'(x)=\sqrt{x+3}+\frac{x}{ 2 \sqrt{x+3}} \\ f'(x) =\frac{ \sqrt{x+3} }{1} \cdot \frac{2 \sqrt{x+3}}{2 \sqrt{x+3}} +\frac{x}{2 \sqrt{x+3}} \\ f'(x)=\frac{2(\sqrt{x+3})^2+x}{2 \sqrt{x+3}}\]

- anonymous

So is it x=-2 for when f'(x) = 0

- freckles

yes

- anonymous

and then the denominator would give me when derivative is undefined?

- anonymous

so the critical points are x= -2, and -3?

- freckles

in by the way the domain of f is x>=-3
so we have this so far:
|dw:1437451331355:dw|

- freckles

x=-3 is an endpoint

- freckles

since the function doesn't exist to the left of x=-3

- freckles

so you only have two intervals to consider
the ones in the picture above
just choose a number from each
and plug into f'(x)

- UsukiDoll

what about the interval (-oo, -3) . Don't we have to pick a number in that interval too?

- freckles

the function doesn't exist to the left of x=-3

- freckles

the domain of x *sqrt(x+3)
is x+3>=0 because of that square root function

- UsukiDoll

yeah and we can't have x = -3.
I can see that part, but how did you guys get x = -2 ?

- freckles

well we can have x=-3 but we can't have anything less than x=-3
and he and she and I got x=-2 from finding the critical numbers

- freckles

The critical numbers are the numbers that satisfy either f'(x)=0 or f'(x) does not exist as long as those x values are in the domain of the original function f(x).

- anonymous

why cant we say the interval is from (-3, Infinity)

- freckles

remember @mary_am you found x=-2 as a critical number
so we have to consider the intervals (-3,-2) and (-2,infty) separately

- anonymous

gotcha, I think im overthinking it

- freckles

choose a number from each of the intervals (not an endpoint)

- freckles

plug into f'(x)

- anonymous

so its increasing not decreasing on those two diff intervals, right?

- UsukiDoll

\[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0 \]
\[\frac{ 3x+6}{2 \sqrt{x+3}}=0\]
\[\frac{ 3(x+2)}{2 \sqrt{x+3}}=0\]
x=-3, -2 our intervals are based on the critical points
(-3,-2) and (-2,oo)
then pick a number between those intervals and plug it into f'(x)

- anonymous

ya plugged the numbers in and got increasing for both those intervals, but just wanted to make sure I was correct

- UsukiDoll

if it's negative <- decreasing
if it's positive <- increasing

- anonymous

and also would there be any local max or min values for this problem?

- anonymous

because if its just increasing....I would believe that x=-3 would just be the local min...? but no so sure actually

- UsukiDoll

we'll you were asked to find just increasing and decreasing so finding local max and local max isn't required.

- UsukiDoll

which is the first derivative test

- anonymous

lol the second part of the questions asks for the local max and min

- UsukiDoll

I don't even see that you've typed that.

- Zale101

You can find local min and max after you find what intervals are decreasing or increasing.

- anonymous

yea I just did that and I got increasing for both invtervals and now im like super confused

- Zale101

|dw:1437451995489:dw|
If the graph goes from increasing to decreasing, then that point can be a local max, if it goes from decreasing to increasing then it is local min.

- freckles

\[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ \text{ a number from } (-2,-3) \text{ is } -2.5 \\ f'(-2.5)=\frac{3(-2.5)+6}{2 \sqrt{-2.5+3}}=\frac{-6.5+6}{2 \sqrt{.5}}\]
is this positive or negative @mary_am ?

- anonymous

should be negative

- freckles

oops I meant (-3,-2)

- freckles

but yes
f'<0 on (-3,-2)

- freckles

f'<0 means f' is negative

- freckles

if f' is negative on (-3,-2) then f is increasing/decreasing on (-3,-2)?

- anonymous

Ifound my mistake, I plugged in 0...

- freckles

yeah 0 cannot be chosen from (-3,-2)

- anonymous

lol yea....

- UsukiDoll

for (-3,-2) we can only choose a very small selection

- UsukiDoll

x = 0 is more appropriate in (-2,oo) though

- freckles

you could choose some of the following:
-2.99
-2.88
-2.8
-2.7
-2.6
-2.4
-2.1
any many more...

- anonymous

okay, so its increasing from -2, infitnity and decreasing from (-3,-2), right?

- anonymous

Yea I know I just dont know what happened... brain isnt functioning to properly

- UsukiDoll

mine isn't either . It's in force to be awake mode.

- freckles

ok sounds great to me
except I would say on not from
increasing on (-2,infty)
decreasing on (-3,-2)

- freckles

|dw:1437452317669:dw|
so the function may look sorta like this
I just used the information from f' and the domain of the original function to help me draw a rough graph of f

- freckles

|dw:1437452372080:dw|

- freckles

reading the graph from left to right of course

- anonymous

But I am right when I say that the fucntion inreases from -2 to infinity and decreases from -3 to -2, right? and that there is a local minumum at x=-2

- freckles

yes
but that local min can also be a ?

- UsukiDoll

it's backwards
I would write, our function is decreasing from (-3,-2) and increasing from (-2,oo)

- anonymous

OH it can also be an inflection point....?

- freckles

Well I was going for global max
inflection point you might want to consider when f''=0

- anonymous

I also have to find the intervals for cancavity and inflection points

- freckles

global min*

- anonymous

so I know that Im supposed to the second derivative

- anonymous

I've never heard of global min...

- freckles

|dw:1437452577062:dw|

- freckles

it is just the lowest point on the graph

- UsukiDoll

for concavity take the derivative again... set that second derivative to 0 and find the critical points

- anonymous

So is the second derivative = (x+3^-1/2) (2) - (2x+3)(-1/2x+3^-3/2)

- anonymous

or am I like waaaay off....? :/

- freckles

\[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ f'(x)=\frac{(3x+6)' \cdot 2 \sqrt{x+3}-(2 \sqrt{x+3})' \cdot (3x+6)}{(2 \sqrt{x+3})^2}\]

- freckles

can you find (3x+6)'
and (2 sqrt(x+3))'

- anonymous

yea, hollup

- anonymous

its 3 and the other one is x+3 ^-1/2

- freckles

that is right \[(3x+6)'=(3x)'+(6)'=3+0=3 \\ (2 \sqrt{x+3})'=2 (\sqrt{x+3})'=2 \frac{1}{2 \sqrt{x+3}} = \frac{1}{\sqrt{x+3}}\]

- freckles

\[f''(x)=\frac{3 \cdot 2 \sqrt{x+3}-\frac{1 }{\sqrt{x+3}} \cdot (3x+6)}{4(x+3)}\]

- freckles

you multiply bottom and top by sqrt(x+3) to clear the compound fraction

- anonymous

\[(2\sqrt{X+3) (3) - }\]

- anonymous

what the heck...why didnt my answer show up

- anonymous

okay ill do what you just asked

- anonymous

so I got 6(sqrt x+3) * (sqrt x+3) - (3x-6)(sqrt x+3) all over 4(x+3)*(sqrt x+3)

- anonymous

should I simplify more?

- freckles

\[f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}\]

- anonymous

yea, basicaly the same thing!

- UsukiDoll

now we need critical points.

- freckles

except you have a sqrt(x+3) next to the (3x+6)

- anonymous

yea thanks for pointing that out!

- freckles

anyways do you know where to go from here?

- anonymous

so x= -4 is one of the critical points

- freckles

find the POSSIBLE inflection points by finding where f''=0
f'' does not exist at x=-3 we already know at x=-3 we do not have an inflection point since it is an endpoint

- freckles

\[6(x+3)-3x-6=0 \\ 3x+12=0 \\ x=-4\]
yeah sounds great

- freckles

now to find if (-4,f(-4)) is an inflection points test the intervals to see if the concavity switches

- anonymous

and for the undefined crtical point its going to x= -3?

- anonymous

Do I also put x= -2 on the number line when testing out the points?

- anonymous

|dw:1437454017504:dw|

- anonymous

and then I test points in between those numbers with the second derivative to find where its CU and CD?

- freckles

you have one possible inflection point
so you only have two intervals to check

- anonymous

so just stick with -3 and -2 ?

- anonymous

okay

- freckles

lol

- freckles

wait sorry x=-4 wasn't on (-3,infty) which is the domain of f

- freckles

we don't have an inflection point

- freckles

x=-4 wasn't in the domain of the original function

- anonymous

thats super odd

- freckles

the function is either concave up or concave down

- freckles

it doesn't have a combination of the two

- anonymous

so the answer is like nothing

- freckles

test any number in the domain of f
that is test any number from (-3,inf)
to see which it is

- anonymous

I plugged in -3.5 to test it out between the intervals of (-3,-4)

- freckles

put that number into f''

- freckles

(-3,-4) doesn't make sense
because -4 isn't greater than -3

- anonymous

-4,-3

- UsukiDoll

the bigger the negative number, the smaller its values is

- freckles

but (-4,-3) isn't in the domain of f

- anonymous

kk lemme plug in -2 then

- freckles

you are forgetting the function has domain (-3,infty)

- UsukiDoll

is it because at x =-3 the whole function is undefined?

- freckles

you can choose a number from (-3,infty) to test the concavity of f
by using f''

- anonymous

so when I used -2 I got a negative number out which was -1.2

- freckles

\[f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}\]
\[f''(x)=\frac{3x+12}{4(x+3)\sqrt{x+3} } \\ f''(0)=\frac{3(0)+12}{4(0+3)\sqrt{0+3}} \\\ f''(0)=\frac{12}{4(3)\sqrt{3}}>0 \\ \text{ hmm... you got a negative number \let me see } \\ f''(-2)=\frac{3(-2)+12}{4(-2+3)\sqrt{-2+3}}=\frac{6}{4(1)\sqrt{1}}\]
don't see how you got a negative number :(

- anonymous

wow thats odd, I plugged the number into the first equation you put above but that shouldnt have given me a diff answer

- freckles

you mean the unsimplfied version I have of f'' ?

- anonymous

basicalllly

- anonymous

okay i did it again and i got 6/4

- anonymous

so this is how my number line looks like

- freckles

how does it look?

- anonymous

|dw:1437454970480:dw|

- freckles

right because our function only exists to the right of and including x=-3

- anonymous

sorry for the ratchetness of it, still getting used to it

- freckles

what is all the + signs mean
are you just saying it is concave up on (-3,infty)?

- anonymous

ya

- freckles

ok cool

- anonymous

so is there also an inflection point at x=-3?

- freckles

anyways I must leave you
it is passed my bed time

- freckles

no there is no inflection point

- freckles

the concavity does not switch ever

- nincompoop

laughing out loud

- freckles

the function is concave up everywhere on the function's domain

- anonymous

thanks a lot, you really are a great person! Bless yo soul dude

- anonymous

LOL I really wish you were my calc teacher instead

- freckles

it was no problem
peace for now

- anonymous

Peace freckles

- nincompoop

ye a teacher that solves your own homework! that would be nice

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