anonymous
  • anonymous
NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3) * This is a calc problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
Have you considered finding the derivative?
UsukiDoll
  • UsukiDoll
we need the first derivative test... wait we need to find critical points first!
freckles
  • 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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UsukiDoll
  • UsukiDoll
what @freckles said is correct... use product rule
anonymous
  • 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
  • freckles
I don't think we need to find the x-intercepts to find where the function is increasing/decreasing.
anonymous
  • anonymous
is it just x= -3?
UsukiDoll
  • UsukiDoll
noooooo....................... freckles we need the critical points first or we can't go further T_T
UsukiDoll
  • UsukiDoll
sorry mary that comment was at freckles
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
  • Zale101
To get the critical points, we first need to set the derivative to zero
anonymous
  • anonymous
Yea I did that already guys!!!
Zale101
  • Zale101
and solve for x. Those x's are the crtical points.
freckles
  • freckles
good then what did you for the x's?
anonymous
  • anonymous
I just need help in solving for x
freckles
  • freckles
\[\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}\] find common denominator
anonymous
  • anonymous
No, we find the first derivative and then find the critical points
freckles
  • freckles
\[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0\]
anonymous
  • anonymous
hey freckles, how did you get +x in the numerator?
freckles
  • freckles
(sqrt x+3) + x/(2(sqrt x +3) isn't this what you got for f'?
anonymous
  • anonymous
OH, K i see it
freckles
  • 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
  • anonymous
So is it x=-2 for when f'(x) = 0
freckles
  • freckles
yes
anonymous
  • anonymous
and then the denominator would give me when derivative is undefined?
anonymous
  • anonymous
so the critical points are x= -2, and -3?
freckles
  • freckles
in by the way the domain of f is x>=-3 so we have this so far: |dw:1437451331355:dw|
freckles
  • freckles
x=-3 is an endpoint
freckles
  • freckles
since the function doesn't exist to the left of x=-3
freckles
  • 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
  • UsukiDoll
what about the interval (-oo, -3) . Don't we have to pick a number in that interval too?
freckles
  • freckles
the function doesn't exist to the left of x=-3
freckles
  • freckles
the domain of x *sqrt(x+3) is x+3>=0 because of that square root function
UsukiDoll
  • UsukiDoll
yeah and we can't have x = -3. I can see that part, but how did you guys get x = -2 ?
freckles
  • 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
  • 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
  • anonymous
why cant we say the interval is from (-3, Infinity)
freckles
  • 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
  • anonymous
gotcha, I think im overthinking it
freckles
  • freckles
choose a number from each of the intervals (not an endpoint)
freckles
  • freckles
plug into f'(x)
anonymous
  • anonymous
so its increasing not decreasing on those two diff intervals, right?
UsukiDoll
  • 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
  • anonymous
ya plugged the numbers in and got increasing for both those intervals, but just wanted to make sure I was correct
UsukiDoll
  • UsukiDoll
if it's negative <- decreasing if it's positive <- increasing
anonymous
  • anonymous
and also would there be any local max or min values for this problem?
anonymous
  • anonymous
because if its just increasing....I would believe that x=-3 would just be the local min...? but no so sure actually
UsukiDoll
  • UsukiDoll
we'll you were asked to find just increasing and decreasing so finding local max and local max isn't required.
UsukiDoll
  • UsukiDoll
which is the first derivative test
anonymous
  • anonymous
lol the second part of the questions asks for the local max and min
UsukiDoll
  • UsukiDoll
I don't even see that you've typed that.
Zale101
  • Zale101
You can find local min and max after you find what intervals are decreasing or increasing.
anonymous
  • anonymous
yea I just did that and I got increasing for both invtervals and now im like super confused
Zale101
  • 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
  • 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
  • anonymous
should be negative
freckles
  • freckles
oops I meant (-3,-2)
freckles
  • freckles
but yes f'<0 on (-3,-2)
freckles
  • freckles
f'<0 means f' is negative
freckles
  • freckles
if f' is negative on (-3,-2) then f is increasing/decreasing on (-3,-2)?
anonymous
  • anonymous
Ifound my mistake, I plugged in 0...
freckles
  • freckles
yeah 0 cannot be chosen from (-3,-2)
anonymous
  • anonymous
lol yea....
UsukiDoll
  • UsukiDoll
for (-3,-2) we can only choose a very small selection
UsukiDoll
  • UsukiDoll
x = 0 is more appropriate in (-2,oo) though
freckles
  • 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
  • anonymous
okay, so its increasing from -2, infitnity and decreasing from (-3,-2), right?
anonymous
  • anonymous
Yea I know I just dont know what happened... brain isnt functioning to properly
UsukiDoll
  • UsukiDoll
mine isn't either . It's in force to be awake mode.
freckles
  • freckles
ok sounds great to me except I would say on not from increasing on (-2,infty) decreasing on (-3,-2)
freckles
  • 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
  • freckles
|dw:1437452372080:dw|
freckles
  • freckles
reading the graph from left to right of course
anonymous
  • 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
  • freckles
yes but that local min can also be a ?
UsukiDoll
  • UsukiDoll
it's backwards I would write, our function is decreasing from (-3,-2) and increasing from (-2,oo)
anonymous
  • anonymous
OH it can also be an inflection point....?
freckles
  • freckles
Well I was going for global max inflection point you might want to consider when f''=0
anonymous
  • anonymous
I also have to find the intervals for cancavity and inflection points
freckles
  • freckles
global min*
anonymous
  • anonymous
so I know that Im supposed to the second derivative
anonymous
  • anonymous
I've never heard of global min...
freckles
  • freckles
|dw:1437452577062:dw|
freckles
  • freckles
it is just the lowest point on the graph
UsukiDoll
  • UsukiDoll
for concavity take the derivative again... set that second derivative to 0 and find the critical points
anonymous
  • anonymous
So is the second derivative = (x+3^-1/2) (2) - (2x+3)(-1/2x+3^-3/2)
anonymous
  • anonymous
or am I like waaaay off....? :/
freckles
  • 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
  • freckles
can you find (3x+6)' and (2 sqrt(x+3))'
anonymous
  • anonymous
yea, hollup
anonymous
  • anonymous
its 3 and the other one is x+3 ^-1/2
freckles
  • 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
  • freckles
\[f''(x)=\frac{3 \cdot 2 \sqrt{x+3}-\frac{1 }{\sqrt{x+3}} \cdot (3x+6)}{4(x+3)}\]
freckles
  • freckles
you multiply bottom and top by sqrt(x+3) to clear the compound fraction
anonymous
  • anonymous
\[(2\sqrt{X+3) (3) - }\]
anonymous
  • anonymous
what the heck...why didnt my answer show up
anonymous
  • anonymous
okay ill do what you just asked
anonymous
  • 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
  • anonymous
should I simplify more?
freckles
  • freckles
\[f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}\]
anonymous
  • anonymous
yea, basicaly the same thing!
UsukiDoll
  • UsukiDoll
now we need critical points.
freckles
  • freckles
except you have a sqrt(x+3) next to the (3x+6)
anonymous
  • anonymous
yea thanks for pointing that out!
freckles
  • freckles
anyways do you know where to go from here?
anonymous
  • anonymous
so x= -4 is one of the critical points
freckles
  • 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
  • freckles
\[6(x+3)-3x-6=0 \\ 3x+12=0 \\ x=-4\] yeah sounds great
freckles
  • freckles
now to find if (-4,f(-4)) is an inflection points test the intervals to see if the concavity switches
anonymous
  • anonymous
and for the undefined crtical point its going to x= -3?
anonymous
  • anonymous
Do I also put x= -2 on the number line when testing out the points?
anonymous
  • anonymous
|dw:1437454017504:dw|
anonymous
  • anonymous
and then I test points in between those numbers with the second derivative to find where its CU and CD?
freckles
  • freckles
you have one possible inflection point so you only have two intervals to check
anonymous
  • anonymous
so just stick with -3 and -2 ?
anonymous
  • anonymous
okay
freckles
  • freckles
lol
freckles
  • freckles
wait sorry x=-4 wasn't on (-3,infty) which is the domain of f
freckles
  • freckles
we don't have an inflection point
freckles
  • freckles
x=-4 wasn't in the domain of the original function
anonymous
  • anonymous
thats super odd
freckles
  • freckles
the function is either concave up or concave down
freckles
  • freckles
it doesn't have a combination of the two
anonymous
  • anonymous
so the answer is like nothing
freckles
  • freckles
test any number in the domain of f that is test any number from (-3,inf) to see which it is
anonymous
  • anonymous
I plugged in -3.5 to test it out between the intervals of (-3,-4)
freckles
  • freckles
put that number into f''
freckles
  • freckles
(-3,-4) doesn't make sense because -4 isn't greater than -3
anonymous
  • anonymous
-4,-3
UsukiDoll
  • UsukiDoll
the bigger the negative number, the smaller its values is
freckles
  • freckles
but (-4,-3) isn't in the domain of f
anonymous
  • anonymous
kk lemme plug in -2 then
freckles
  • freckles
you are forgetting the function has domain (-3,infty)
UsukiDoll
  • UsukiDoll
is it because at x =-3 the whole function is undefined?
freckles
  • freckles
you can choose a number from (-3,infty) to test the concavity of f by using f''
anonymous
  • anonymous
so when I used -2 I got a negative number out which was -1.2
freckles
  • 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
  • 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
  • freckles
you mean the unsimplfied version I have of f'' ?
anonymous
  • anonymous
basicalllly
anonymous
  • anonymous
okay i did it again and i got 6/4
anonymous
  • anonymous
so this is how my number line looks like
freckles
  • freckles
how does it look?
anonymous
  • anonymous
|dw:1437454970480:dw|
freckles
  • freckles
right because our function only exists to the right of and including x=-3
anonymous
  • anonymous
sorry for the ratchetness of it, still getting used to it
freckles
  • freckles
what is all the + signs mean are you just saying it is concave up on (-3,infty)?
anonymous
  • anonymous
ya
freckles
  • freckles
ok cool
anonymous
  • anonymous
so is there also an inflection point at x=-3?
freckles
  • freckles
anyways I must leave you it is passed my bed time
freckles
  • freckles
no there is no inflection point
freckles
  • freckles
the concavity does not switch ever
nincompoop
  • nincompoop
laughing out loud
freckles
  • freckles
the function is concave up everywhere on the function's domain
anonymous
  • anonymous
thanks a lot, you really are a great person! Bless yo soul dude
anonymous
  • anonymous
LOL I really wish you were my calc teacher instead
freckles
  • freckles
it was no problem peace for now
anonymous
  • anonymous
Peace freckles
nincompoop
  • nincompoop
ye a teacher that solves your own homework! that would be nice

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