#Functions problem
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You're on the right track. Now, recall the range of arcsin(x).
minus pi/2 to pi/2
Right.
Well, the first inequality isn't quite right: -π/2 is excluded, since it would make the expression under the logarithm zero.
right.. so a normal paranthesis then
On which side?
Yes.
like this?
this is done in my book also... but the thing im confused about is
we calculated the domain with ONLY the inequality given by range of arcsin....
we should consider that [pi + 2arcsin{ 3 - x /7 }]/pi greater than 0
because log wouldnt be defined in if we dont check that... we literally ignored that one
You literally started with that inequality, though...
yeah but later on we solved it thru that range of arcsin
Yes.
we didnt take intersection
Intersection of what?
Oh, I see what you mean.
That isn't needed when dealing with a composite function.
As in, that happens "automatically", so to speak.
composite function?
Yeah, like here.
But if we had a sum or a product, then yes, we'd have to deal with them separately and intersect their domains.
whats that?
What's what?
Oh.
Well, a composition of f(x) and g(x) is f(g(x)).
Because the inequality we took is how the domain of a logarithmic function is found.
okay so if g{x} is defined then f{x} will automatically defined?
Not sure what you mean.
We have a function of the form ln(f(x)). The condition to find its domain is just f(x) > 0.
did we do that? like we just directly put the range of f{x} and got that inequality of x...
Yes, that's the inequality you wrote.
does that mean if there is a function f{x} inside log
then we dont need to check for which values log will be defined and can directly find domain of the f{x}?
I'm not sure what exactly you're asking.
Logarithmic functions are defined when the expression under the logarithm is positive. That's it.
If we have a function of the form ln(f(x)), then you can just find its domain by solving f(x) > 0.
um... I started it fresh... and directly got the answer from range of arcsin...
I didn't even have to consider that entire thing being greater than 0
didn't even have to solve this
oh wait wait... this gives me the minimum value of arcsin for this function no?
like let that 3-x/7 = y
so arcsiny > -pi/2 .... {i}
and we know range is
-pi/2 <= arcsin y <= pi/2 ...... {ii}
taking intersection of both we get
-pi/2 < arcsin y <= pi/2 ....... {iii}
and solving {iii} gives us domain
If you don't do that, then you get the domain as [-4, 10], which is incorrect.
i think i get it..
lets say we have f[g{x}]
we can get domain of g{x} and then eliminate all values for f[g{x}] to be undefined... and finally we get the domain
Well, you can do that, yes. It's pretty much the same process, but not as straightforward.
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Remember to find the range, too.
yessir... its easy just will put range of arcsin
then multiply it with 2, followed by -1, inequalities flipped, add pi, then divide by pi. and then take log of the entire thing
will get something like minus infinity ig so will put paranthesis on that side