Question

The expression dy/dx = x(3√y) gives the slope at any point on the graph of the function f(x) where f(2) = 8.

Answer 1

B. Write an expression for f(x) in terms of x.
C.What is the domain of f(x)?
D. What is the minimum value of f(x)?
E. Using the axes provided, sketch a slope field for the given differential equation at the nine points indicated.

Answer 2

separate the variables
then integrate

Answer 3

Is the first step to divide by x to get 3sqrt(y) alone?

Answer 4

you should have this:
\[\frac{1}{3 \sqrt{y}} dy=xdx\]
now just integrate both sides

Answer 5

I got 2sqrt(y)/3 + C = x^2/2 + C

Answer 6

Is the + constant on both sides necessary?

Answer 7

no a constant - another constant or + another constant is still a constant

Answer 8

just to be sure your problem was
\[\frac{dy}{dx}=x (3 \sqrt{y}) \]right?

Answer 9

if so your answer is good so far

Answer 10

Yes that was the problem. Also, just to be sure, I do or don't need the constant?

Answer 11

no a constant - another constant or + another constant is still a constant

for example C-k is still a constant and you can call it K

Answer 12

you just need one constant on one side

Answer 13

Alright thanks!

Answer 14

Now how do I do part C?

Answer 15

use the condition given

Answer 16

do you see that f(2)=8 thing?

Answer 17

Yes,

Answer 18

@freckles

Answer 19

yep? having trouble?

Answer 20

Well I know I need to plug something in for y, but I'm not sure what.

Answer 21

you have the condition f(2)=8
or you can say y(2)=8
so to find the C you do:
this means to replace x with 2
and y with 8

Answer 22

so 2sqrt(8)/3 + C= 2^2/2

Answer 23

Your C is going to look pretty ugly which is fine...
but I want to ask one more time
the problem is:
\[\frac{dy}{dx}=x(3 \sqrt{y}) ?\]

and not something like:
\[\frac{dy}{dx}=x(\sqrt[3]{y}) ?\]

Answer 24

The first one.

Answer 25

K well you can continue solving for C
then you have your f(x) that satisfies the condition given

Answer 26

I got C= 4sqrt(2)/3 -2

Answer 27

can I see how you solved for C?

Answer 28

which side did you put C on ?
I thought you had it on the left hand side with the y at first
it seems you reversed it

Answer 29

Sorry I got C= - 4sqrt(2)/3 +2

Answer 30

ok sounds better

Answer 31

So would I have that as my final answer or do I plug it in somewhere?

Answer 32

you already integrated
replace the C in your solution with the C you just found

Answer 33

you will also need to solve for y since they are asking for an explicit form for f(x)

Answer 34

2sqrt(y)/3 - 4sqrt(2)/3 +2 = x^2/2

Answer 35

yeah we might need to keep in mind that y>0 for later
so we solve that for y

Answer 36

I got y > or equal to 0

Answer 37

ok well solve for y

Answer 38

When I solved for y I got 1/16 (272-192 sqrt(2)-72 x^2+48 sqrt(2) x^2+9 x^4)

Answer 39

I think I would leave it in pretty form
I don't know about expanding all that :p
\[2 \frac{\sqrt{y}}{3}-\frac{4 \sqrt{2}}{3}+2=\frac{x^2}{2} \\ \frac{2}{3} \sqrt{y}=\frac{x^2}{2}+\frac{4 \sqrt{2}}{3}-2 \\ \frac{2}{3} \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{3}{2} \cdot \frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{\color{red}{\cancel{3}}}{2} \frac{3x^2+8 \sqrt{2}-12}{ \color{red}{\cancel{(3)}}2} \\ \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{4} \\ y \ge 0 \\ y=(\frac{3x^2+8 \sqrt{2}-12}{4})^2 \]
but I don't think there are any restrictions on x...
since for all x the right hand side is positive or zero as y should be

Answer 40

but I could be wrong

Answer 41

So in the end. The domain of f(x) is y > or equal to 0?

Answer 42

I did something way more complicated, but I got the same answer.

Answer 43

well you are thinking range when you state domain just now

Answer 44

I mean you stated the range just now

Answer 45

Sorry i meant the domain is all real numbers. The range is y > or equal to 0.

Answer 46

yep that is what I think too

Answer 47

Now how does this lead into D?

Answer 48

you find when y'=0

Answer 49

y'=0 will give us critical numbers

Answer 50

Does that me I do d/dx (3x^2+8sqrt(2)-12/4)^2

Answer 51

you are already have y' above:

just substitute what we have for sqrt(y) above into it in terms of x

Answer 52

So 2sqrt(x)/3 -4sqrt(2)/3 +2 = 0

Answer 53

I have a problem with our domain

Answer 54

recall this:
\[2 \frac{\sqrt{y}}{3}-\frac{4 \sqrt{2}}{3}+2=\frac{x^2}{2} \\ \frac{2}{3} \sqrt{y}=\frac{x^2}{2}+\frac{4 \sqrt{2}}{3}-2 \\ \frac{2}{3} \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{3}{2} \cdot \frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{\color{red}{\cancel{3}}}{2} \frac{3x^2+8 \sqrt{2}-12}{ \color{red}{\cancel{(3)}}2} \\ \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{4} \\ y \ge 0 \\ y=(\frac{3x^2+8 \sqrt{2}-12}{4})^2\]
look at my 4th row before solving for y

Answer 55

sqrt(y) should only output 0 or positive
is there any x that would make the right hand side negative?

Answer 56

No?

Answer 57

the numerator needs to be >=0

Answer 58

Oh, my bad.

Answer 59

So does that mean the minimum value of f(x) is 0?

Answer 60

no
I said I had a problem with our domain

Answer 61

As in it's wrong?

Answer 62

\[2 \frac{\sqrt{y}}{3}-\frac{4 \sqrt{2}}{3}+2=\frac{x^2}{2} \\ \frac{2}{3} \sqrt{y}=\frac{x^2}{2}+\frac{4 \sqrt{2}}{3}-2 \\ \frac{2}{3} \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{3}{2} \cdot \frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{\color{red}{\cancel{3}}}{2} \frac{3x^2+8 \sqrt{2}-12}{ \color{red}{\cancel{(3)}}2} \\ \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{4} \\ y \ge 0 \\ y=(\frac{3x^2+8 \sqrt{2}-12}{4})^2\]
I said the the 4th line here says we need the numerator on the right hand side to be positive or zero

Answer 63

we need
\[3x^2+8 \sqrt{2}-12 \ge 0\]

Answer 64

well you can look at either 4,5,6th line to see this

Answer 65

Does this have to do with question C or D?

Answer 66

I have been talking about the domain
so whichever question asks about domain

Answer 67

C

Answer 68

D ask for the minimum value of f(x).

Answer 69

D is the one I need help on atm.

Answer 70

But I think there is a problem with the domain

Answer 71

you don't want to correct it ?

Answer 72

What needs to be corrected?

Answer 73

have you been reading what I have been saying?
look at the 6th line since it looks more simplified:
\[2 \frac{\sqrt{y}}{3}-\frac{4 \sqrt{2}}{3}+2=\frac{x^2}{2} \\ \frac{2}{3} \sqrt{y}=\frac{x^2}{2}+\frac{4 \sqrt{2}}{3}-2 \\ \frac{2}{3} \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{3}{2} \cdot \frac{3x^2+8 \sqrt{2}-12}{6} \\ \sqrt{y}=\frac{\color{red}{\cancel{3}}}{2} \frac{3x^2+8 \sqrt{2}-12}{ \color{red}{\cancel{(3)}}2} \\ \sqrt{y}=\frac{3x^2+8 \sqrt{2}-12}{4} \\ y \ge 0 \\ y=(\frac{3x^2+8 \sqrt{2}-12}{4})^2\]

sqrt(y) should output 0 or positive numbers
but (3x^2+8sqrt(2)-12))/4 can sometimes be negative

we need (3x^2+8sqrt(2)-12) to be positive

we need to solve:
\[3x^2+8\sqrt{2}-12 \ge 0\]

Answer 74

positive or zero that is

Answer 75

I think the answer would be all real numbers squared?

Answer 76

so you are saying sqrt(y) can output negative numbers?

Answer 77

so since it is already squared, all real numbers?

Answer 78

I cannot think of one real number that I can plug into the sqrt( ) function that will give me a negative number

Answer 79

sqrt(y) say the output is 0 or positive for all real y
therefore 3x^2+8sqrt(2)-12 needs to be 0 or positive

Answer 80

Well I can't think of any x value that can be negative, so doesn't that mean all real numbers?

Answer 81

you need to also look at the equation before we found an explicit from for y

Answer 82

https://www.wolframalpha.com/input/?i=domain+of+2sqrt%28y%29%2F3+-+4sqrt%282%29%2F3+%2B+2+%3D+x%5E2%2F2

Answer 83

that is like saying g(x)=sqrt(x) has range being all real numbers because g^2(x)=x has domain all real numbers

but g(x)=sqrt(x) doesn't have range all real numbers

Answer 84

Did you check my link above @freckles

Answer 85

It says the domain is all real numbers squared, but since the x is already squared in the equation we can just say all real numbers.

Answer 86

No?

Answer 87

no you need to consider the implicit form

Answer 88

https://www.wolframalpha.com/input/?i=sqrt%28y%29%3Dx%5E2-9
here is a good example notice the graph doesn't include anything on the x-interval (-3,3)
because x^2-9 is less than 0 there

Answer 89

You said 3x^2+8sqrt(2)-12))/4 can sometimes be negative, but in your work you had (3x^2+8sqrt(2)-12))/4)^2. Doesn't that mean it's never negative.

Answer 90

https://www.wolframalpha.com/input/?i=sqrt%28y%29%3D%283x%5E2%2B8+sqrt%282%29-12%29%2F4
notice here for our implicit form that the domain excludes some numbers in between something and something else
(those or the something we need to find from the inequality)

Answer 91

you are not considering the the implicit form again

Answer 92

sorry that was a bad example
and totally not what I meant to say because I made type-o

Answer 93

say you have sqrt(x)=y
this is not the same as x=y^2
do you know why?

Answer 94

Well I trust your judgement, I'm just confused on how to find the domain then.

Answer 95

And what my answer should be.

Answer 96

but this other graph is x=y^2


ok say we have it the other way around
sqrt(y)=x
looks like

notice domain here is [0,infty)

but squaring both sides gives us
y=x^2

domain is all real numbers here