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anonymous
 one year ago
NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3)
* This is a calc problem
anonymous
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.2Have you considered finding the derivative?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we need the first derivative test... wait we need to find critical points first!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2f'>0 tells us f is increasing f'<0 tells us f is decreasing Find the critical numbers and consider the domain.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2what @freckles said is correct... use product rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2I don't think we need to find the xintercepts to find where the function is increasing/decreasing.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2noooooo....................... freckles we need the critical points first or we can't go further T_T

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2sorry mary that comment was at freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\sqrt{x+3}+\frac{x}{2 \sqrt{x+3}} \\ \text{ solve } f'(x)=0 \text{ \to find the critical numbers }\]

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0To get the critical points, we first need to set the derivative to zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I did that already guys!!!

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0and solve for x. Those x's are the crtical points.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2good then what did you for the x's?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just need help in solving for x

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}\] find common denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, we find the first derivative and then find the critical points

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey freckles, how did you get +x in the numerator?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2(sqrt x+3) + x/(2(sqrt x +3) isn't this what you got for f'?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\sqrt{x+3}+\frac{x}{ 2 \sqrt{x+3}} \\ f'(x) =\frac{ \sqrt{x+3} }{1} \cdot \frac{2 \sqrt{x+3}}{2 \sqrt{x+3}} +\frac{x}{2 \sqrt{x+3}} \\ f'(x)=\frac{2(\sqrt{x+3})^2+x}{2 \sqrt{x+3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is it x=2 for when f'(x) = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then the denominator would give me when derivative is undefined?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the critical points are x= 2, and 3?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2in by the way the domain of f is x>=3 so we have this so far: dw:1437451331355:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2since the function doesn't exist to the left of x=3

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2what about the interval (oo, 3) . Don't we have to pick a number in that interval too?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the function doesn't exist to the left of x=3

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the domain of x *sqrt(x+3) is x+3>=0 because of that square root function

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah and we can't have x = 3. I can see that part, but how did you guys get x = 2 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0why cant we say the interval is from (3, Infinity)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2remember @mary_am you found x=2 as a critical number so we have to consider the intervals (3,2) and (2,infty) separately

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0gotcha, I think im overthinking it

freckles
 one year ago
Best ResponseYou've already chosen the best response.2choose a number from each of the intervals (not an endpoint)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its increasing not decreasing on those two diff intervals, right?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0 \] \[\frac{ 3x+6}{2 \sqrt{x+3}}=0\] \[\frac{ 3(x+2)}{2 \sqrt{x+3}}=0\] 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
 one year ago
Best ResponseYou've already chosen the best response.0ya plugged the numbers in and got increasing for both those intervals, but just wanted to make sure I was correct

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2if it's negative < decreasing if it's positive < increasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also would there be any local max or min values for this problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because if its just increasing....I would believe that x=3 would just be the local min...? but no so sure actually

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we'll you were asked to find just increasing and decreasing so finding local max and local max isn't required.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2which is the first derivative test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol the second part of the questions asks for the local max and min

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I don't even see that you've typed that.

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0You can find local min and max after you find what intervals are decreasing or increasing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea I just did that and I got increasing for both invtervals and now im like super confused

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0dw: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
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ \text{ a number from } (2,3) \text{ is } 2.5 \\ f'(2.5)=\frac{3(2.5)+6}{2 \sqrt{2.5+3}}=\frac{6.5+6}{2 \sqrt{.5}}\] is this positive or negative @mary_am ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but yes f'<0 on (3,2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2f'<0 means f' is negative

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if f' is negative on (3,2) then f is increasing/decreasing on (3,2)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ifound my mistake, I plugged in 0...

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yeah 0 cannot be chosen from (3,2)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2for (3,2) we can only choose a very small selection

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2x = 0 is more appropriate in (2,oo) though

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you could choose some of the following: 2.99 2.88 2.8 2.7 2.6 2.4 2.1 any many more...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, so its increasing from 2, infitnity and decreasing from (3,2), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I know I just dont know what happened... brain isnt functioning to properly

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2mine isn't either . It's in force to be awake mode.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok sounds great to me except I would say on not from increasing on (2,infty) decreasing on (3,2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw: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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437452372080:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2reading the graph from left to right of course

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.2yes but that local min can also be a ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it's backwards I would write, our function is decreasing from (3,2) and increasing from (2,oo)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH it can also be an inflection point....?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Well I was going for global max inflection point you might want to consider when f''=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I also have to find the intervals for cancavity and inflection points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I know that Im supposed to the second derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've never heard of global min...

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437452577062:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it is just the lowest point on the graph

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2for concavity take the derivative again... set that second derivative to 0 and find the critical points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is the second derivative = (x+3^1/2) (2)  (2x+3)(1/2x+3^3/2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or am I like waaaay off....? :/

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ f'(x)=\frac{(3x+6)' \cdot 2 \sqrt{x+3}(2 \sqrt{x+3})' \cdot (3x+6)}{(2 \sqrt{x+3})^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2can you find (3x+6)' and (2 sqrt(x+3))'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its 3 and the other one is x+3 ^1/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that 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
 one year ago
Best ResponseYou've already chosen the best response.2\[f''(x)=\frac{3 \cdot 2 \sqrt{x+3}\frac{1 }{\sqrt{x+3}} \cdot (3x+6)}{4(x+3)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you multiply bottom and top by sqrt(x+3) to clear the compound fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(2\sqrt{X+3) (3)  }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what the heck...why didnt my answer show up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay ill do what you just asked

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I got 6(sqrt x+3) * (sqrt x+3)  (3x6)(sqrt x+3) all over 4(x+3)*(sqrt x+3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should I simplify more?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f''(x)=\frac{6(x+3)(3x+6)}{4 (x+3) \sqrt{x+3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea, basicaly the same thing!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now we need critical points.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2except you have a sqrt(x+3) next to the (3x+6)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea thanks for pointing that out!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2anyways do you know where to go from here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so x= 4 is one of the critical points

freckles
 one year ago
Best ResponseYou've already chosen the best response.2find 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
 one year ago
Best ResponseYou've already chosen the best response.2\[6(x+3)3x6=0 \\ 3x+12=0 \\ x=4\] yeah sounds great

freckles
 one year ago
Best ResponseYou've already chosen the best response.2now to find if (4,f(4)) is an inflection points test the intervals to see if the concavity switches

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and for the undefined crtical point its going to x= 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I also put x= 2 on the number line when testing out the points?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437454017504:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then I test points in between those numbers with the second derivative to find where its CU and CD?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you have one possible inflection point so you only have two intervals to check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so just stick with 3 and 2 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2wait sorry x=4 wasn't on (3,infty) which is the domain of f

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we don't have an inflection point

freckles
 one year ago
Best ResponseYou've already chosen the best response.2x=4 wasn't in the domain of the original function

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the function is either concave up or concave down

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it doesn't have a combination of the two

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is like nothing

freckles
 one year ago
Best ResponseYou've already chosen the best response.2test any number in the domain of f that is test any number from (3,inf) to see which it is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I plugged in 3.5 to test it out between the intervals of (3,4)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2put that number into f''

freckles
 one year ago
Best ResponseYou've already chosen the best response.2(3,4) doesn't make sense because 4 isn't greater than 3

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the bigger the negative number, the smaller its values is

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but (4,3) isn't in the domain of f

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0kk lemme plug in 2 then

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you are forgetting the function has domain (3,infty)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2is it because at x =3 the whole function is undefined?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you can choose a number from (3,infty) to test the concavity of f by using f''

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when I used 2 I got a negative number out which was 1.2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f''(x)=\frac{6(x+3)(3x+6)}{4 (x+3) \sqrt{x+3}}\] \[f''(x)=\frac{3x+12}{4(x+3)\sqrt{x+3} } \\ f''(0)=\frac{3(0)+12}{4(0+3)\sqrt{0+3}} \\\ f''(0)=\frac{12}{4(3)\sqrt{3}}>0 \\ \text{ hmm... you got a negative number \let me see } \\ f''(2)=\frac{3(2)+12}{4(2+3)\sqrt{2+3}}=\frac{6}{4(1)\sqrt{1}}\] don't see how you got a negative number :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow thats odd, I plugged the number into the first equation you put above but that shouldnt have given me a diff answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you mean the unsimplfied version I have of f'' ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i did it again and i got 6/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this is how my number line looks like

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437454970480:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2right because our function only exists to the right of and including x=3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry for the ratchetness of it, still getting used to it

freckles
 one year ago
Best ResponseYou've already chosen the best response.2what is all the + signs mean are you just saying it is concave up on (3,infty)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is there also an inflection point at x=3?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2anyways I must leave you it is passed my bed time

freckles
 one year ago
Best ResponseYou've already chosen the best response.2no there is no inflection point

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the concavity does not switch ever

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the function is concave up everywhere on the function's domain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks a lot, you really are a great person! Bless yo soul dude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0LOL I really wish you were my calc teacher instead

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it was no problem peace for now

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0ye a teacher that solves your own homework! that would be nice
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