Question

last one:
domain and range of : 2x^2+y^2=1

Answer 1

thought u did one like this earlier

Answer 2

we talked about domain and range, but i got it wrong...

Answer 3

okay, lemme try

Answer 4

domain: solve for y right?

Answer 5

Right..

Answer 6

okay then: \[y ^{2}= -2x ^{2}+1\]
square root both side?

Answer 7

Yep go ahead..

Answer 8

okay \[y=\sqrt{-2x ^{2} +1}\] but thats not right since we cant have negatives under the root

Answer 9

It is right..

Answer 10

But to find domain it is not right..

Answer 11

what will you do to find domain here?

Answer 12

@bronzegoddess I told you the case of square roots..

What we do in case of square roots ??

Answer 13

greater than or equal to zero

Answer 14

Yes then go for it..

Answer 15

so our domain is [1,infinity[ because -2(0)+1= root of 1 which is 1

Answer 16

\[1 - 2x^2 \ge 0 \implies 2x^2 \le 1 \implies x^2 \le \frac{1}{2} \implies x \le \frac{1}{\sqrt{2}}\]

Answer 17

ah i should equate to greater or equal to zero....

Answer 18

If domain will be [1, infinity) then put x = 2

You will get negative that is not right..

Answer 19

Both..

Answer 20

you will equate that to greater as well as equal to..

Answer 21

yes, thats what i meant, so the range will be from our result to infinity?

Answer 22

Find x in term of y..

Answer 23

\[Domain : (-\infty, \frac{1}{\sqrt{2}}]\]

Answer 24

i meant thats our domain?

Answer 25

oh yes because 2 gives a negative

Answer 26

Yes...

Answer 27

Yes..

Answer 28

so i do the same for x=

Answer 29

Yes go ahead...

Answer 30

ok:
\[2x ^{2}= 1-y ^{2}\]
divide by 2: \[x ^{2}= \frac{ 1-y ^{2} }{ 2 }\]
square root:
\[x=\sqrt{\frac{ 1-y ^{2} }{ 2 }}\]
\[\frac{ 1-y ^{2} }{ 2 } \ge 0\]
solve for y:

Answer 31

@waterineyes shouldn't it be -1/sqrt(2)<=x<=1/sqrt(2)?

Answer 32

good so far?

Answer 33

@ghass1978 where?

Answer 34

Yep

My mistake..

Answer 35

good:)

Answer 36

Because if you put x = -3 there you will get again negative there..

Answer 37

am confused

Answer 38

\[x^2 \le \frac{1}{2} \implies x \le \pm \frac{1}{\sqrt{2}}\]

Answer 39

oh ok

Answer 40

This is assigning a domain a range of :

\[[-\frac{1}{\sqrt{2} }, \frac{1}{\sqrt{2}}]\]

Answer 41

ok

Answer 42

Yes go ahead @bronzegoddess you are right till there...

Answer 43

ok ty (:
\[-y ^{2}\ge-1 ----> y ^{2}\ge1 --->y \ge \pm1\]

Answer 44

There is one mistake..

Answer 45

See when you multiply -1 with an inequality then sign gets reversed..

Answer 46

where? the

Answer 47

ah, ok

Answer 48

\[-y^2 \ge -1\]

Answer 49

\[y \le \pm1\]

Answer 50

Here when you multiply -1 then sign reverses..

Yep..

Answer 51

so range is -1 to 1

Answer 52

or 1 to -infinity

Answer 53

See if you have any doubt or confusion then you can verify it..

See you said 1 to - infinity..

Choose one number in this range and plug in for y:

For example I choose -2 :

\[x = \sqrt{\frac{1 - y^2}{2}}\]

Answer 54

oh yeah, thats wrong

Answer 55

\[x = \sqrt{\frac{1 - 4}{2}} = \sqrt{negative}\]

Answer 56

thx for all the patience and help :)

Answer 57

So range will be:

\[Range : [-1, 1]\]

Answer 58

It is okay..

Welcome dear..

And I am a boy..