how to determine largest domain

[unkempt vine]

i dont get how to do both of these exercises , can anyone help me?

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[lofty turtle]

$$ x-\sqrt{y}\in {a>0\mid a\neq 1} $$

[odd lion]

aL

[lofty turtle]

@unkempt vine

[unkempt vine]

thank you , can you also give the reasoning for it? because i would also like to understand it

[untold ember]

@unkempt vine has given 1 rep to @lofty turtle

[lofty turtle]

logarithm is defined for positive numbers

can't divide by zero, so the thing in ln can't be 1

[lofty turtle]

a != 1 i mean

$$ x-\sqrt{y}\in {a>0\mid a\neq 1} $$

@unkempt vine

[odd lion]

aL

[lofty turtle]

so if you plot on the xy plane, you can pick any x,y as long as x-sqrt(y) >0 and is not equal to 1

[unkempt vine]

@lofty turtle thank you , how do i do the second exercise?

[untold ember]

@unkempt vine has given 1 rep to @lofty turtle

[lofty turtle]

$$ -1 < \frac{1}{\ln (x-\sqrt{y})} \leqslant 2 $$

[odd lion]

aL

[lofty turtle]

manipulate this to lose the logarithm and then conclude something about x,y

@unkempt vine

$$ -\ln (x-\sqrt{y})> 1 $$

[odd lion]

aL

[lofty turtle]

for example

[narrow depot]

Oh, this looks interesting.

Let me try.

ln(x - √(y)) ≠ 0
x - √(y) ∈ (0, 1)(1, +∞)
Let's ignore 1 for now.
x - √(y) > 0
x > √(y)
So, the domain should be: x ∈ (√(y), √(y) + 1)⋃(√(y) + 1, +∞), 0 ≤ y < +∞.

As for the second part... Hm.

Maybe level surfaces are worth looking at?

1/ln(x - √(y)) = C
ln(x - √(y)) = 1/C
x - √(y) = e^(1/C)
These are the level curves. We do need to account for the discontinuity at x - √(y) = 1, though. Hm...

[unkempt vine]

thank you, but for the second parti have no clue

[untold ember]

@unkempt vine has given 1 rep to @narrow depot

[narrow depot]

Oh, wait. Why are even trying so hard. We just need to solve the following inequality:
-1 < 1/ln(z) ≤ 2

Well, easy enough. We split this into a system:
ln(z) < -1
ln(z) ≥ 1/2
Then:
0 < z < 1/e
z ≥ √(e)
Thus:
x - √(y) ∈ (0, 1/e)⋃[√(e), +∞)

[unkempt vine]

thank you, what should the sketch look like

[untold ember]

@unkempt vine has given 1 rep to @narrow depot

[narrow depot]

I showed the solution of the inequality below.

[unkempt vine]

i know but i need to draw it , but i don't know what it should loook like

[lofty turtle]

how would you plot if you knew x-sqrt(y) = a @unkempt vine

[unkempt vine]

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y

i think someting like this

[lofty turtle]

what does that even mean

your plot is here

[unkempt vine]

the curves

[lofty turtle]

lets do something simpler for starters

[unkempt vine]

i need to sketch the pre-image

[lofty turtle]

suppose x+y in [0,10]

how do you plot this on xy plane

[unkempt vine]

i have no clue

[lofty turtle]

alright, how do you plot x+y = 1 on xy plane?

[unkempt vine]

well i would change it to y= -x+1 and then draw the line

[lofty turtle]

good

so if you did that for x+y=0 and x+y=10 ..

there's a strip that stays between the two lines

that's the plot

now, let's get back to the initial problem
x-sqrt(y) = sqrt(e)

what does this look like

[unkempt vine]

idk i think you could square the function and then change into y=xm+b or something

[lofty turtle]

so do that

[narrow depot]

Wait, hold up. Why even do that?

[unkempt vine]

well thats familiar for me, but if there is a better way pls explain

[narrow depot]

If we have x - f(y) ∈ (a, b), then that's just the region between x = a + f(y) and x = b + f(y).

[lofty turtle]

that's what I was getting at

[narrow depot]

But why do we need to square anything?

[lofty turtle]

we don't need to

but that's what he proposed and it's not incorrect

[unkempt vine]

ok can you guys explain the steps bc i cant figure out what to draw

[lofty turtle]

plot x = sqrt(y) and x=sqrt(y)+1

[lofty turtle]

zorry x=sqrt(y)+1/e

but the picture is the same

that covers this part of the plot

include this one as well

x=sqrt(y) + sqrt(e)

[lofty turtle]

now your region is made up of the area between (and not including) the first two lines

or the area to the right of the third line, including the line itself

[unkempt vine]

so this is it?

[lofty turtle]

yes

[unkempt vine]

oke thanks

ok

[lofty turtle]

darpinger did most of the heavy lifting, so thank him

[unkempt vine]

both of you

i have one more question

[lofty turtle]

shoot

[unkempt vine]

wait i'll open another post because it has nothing to do with this

[lofty turtle]

sure, close this one if the problem's solved

[unkempt vine]

"
Rock (RUG)
3:53 PM (15 minutes ago)
to me

Your answer has been improved, but there's still room for enhancement. In your attempt, you've depicted three curves on a plane, but what's necessary is a depiction of a set, specifically a shaded region"