Methods 3/4 Questions

[#1procrastinator]

^ Very helpful, thanks a lot. What exactly does well-defined mean? I think I have some idea of what it means but it's not concrete for me

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Other questions

1) In the Essential book, it says for the composition of g with f to be defined, the range of has to be a proper subset of the domain of g. Does this mean that the range has to be a subset of the domain of g but it can't EQUAL the domain of g?

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2) Let a be a positive number, let f: [2, infinity) -> R, f(x) = a-x and g: (-infinity, 1] -> R, g(x) = x^2+a. Find all values of a for which f o g and g o f both exist.

I got the correct answer but not after three tries and I'm looking for a more systemic way of doing it. The way I did it involved a bit of guessing and plugging in different numbers to check.

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3) For f: (-infinity, 2] -> R, f(x) = (x-2)^2

To find the inverse, how do you know to take the negative square root? Is the only way to tell just to look at the domain of f and hopefully realise that it must be the negative root to have as its range?

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4) Why is the horizontal translation in y=2^(-x+2) two units to the right and not left?

[#1procrastinator]

By the way TT (or anyone who can help), with the nested radical problem a few posts ago, is that method with the expression a+sqrt(b) a general way of dealing with such problems or is it a special case? e.g. do you use a+sqrt(b) if you're trying to simplify something in that form?

(different numbers to above)

[nina_rox]

Questions: how do you find maximal domain of square root of (x-4/x+1)?
Thank you!

[#1procrastinator]

Not sure if this is right but here's how I did it (imagine => to be the equal to or greater sign lol)

We require (x-4)/(x+1) be equal to or greater than 0 so then you have either so either both the numerator and the denominator have to be positive or negative, so then if you set the them greater than 0 and less than zero, you get: x=>4 and x>-1 or x <=4 and x < -1

So then the domain would be (-infinity, -1) U [4, infinity)

Someone might want to check that :p

[b^3]

Not sure if this is right but here's how I did it (imagine => to be the equal to or greater sign lol)

We require (x-4)/(x+1) be equal to or greater than 0 so then you have either so either both the numerator and the denominator have to be positive or negative, so then if you set the them greater than 0 and less than zero, you get: x=>4 and x>-1 or x <=4 and x < -1

So then the domain would be (-infinity, -1) U [4, infinity)

Someone might want to check that :p

Looks all good.

[nina_rox]

Not sure if this is right but here's how I did it (imagine => to be the equal to or greater sign lol)

We require (x-4)/(x+1) be equal to or greater than 0 so then you have either so either both the numerator and the denominator have to be positive or negative, so then if you set the them greater than 0 and less than zero, you get: x=>4 and x>-1 or x <=4 and x < -1

So then the domain would be (-infinity, -1) U [4, infinity)

Someone might want to check that :p

Thank you both so much! That makes much more sense now!

[nina_rox]

Another question if you could help:
If f(x) = -x^3 - 2x^2 + x + 3 rewrite so it is in form of (x-a)(b-x)(x+c) + r(x) where r(x) is the remainder?

Thank you!

[kamil9876]

There are infinitely many correct answers to this question. Have you been given any value of a,b,c? (i.e is this just part of some bigger question)

[nina_rox]

Ooopps sorry forgot to post part a) Find the remainder of the function when it is divided by x+2? So I got a remainder of 1. Thanks for your help!

[Phy124]

Ooopps sorry forgot to post part a) Find the remainder of the function when it is divided by x+2? So I got a remainder of 1. Thanks for your help!

We have the equation;



We know part of the equation;



We can expand this and put it in the same form;



Then consequently solve for coefficients;









Giving us;



or



So we have;



or



In which case I would go with the latter.

[nina_rox]

Thank you so much, that makes much more sense!

[#1procrastinator]

How do you rearrange this for x?

ln((x+4)/2) =2e^x-4

[kamil9876]

Just out of curiosity, is this part of some bigger problem? because I don't think there's a nice solution but perhaps you don't need to...

Just one stupid thing I noticed that doesn't help but makes me curious, x is the solution of f(f(x))=x where f is the function f(u)=ln((u+4)/2)

[#1procrastinator]

I'm pretty sure it was just solve for x (probably meant use a CAS) but if it was part of a bigger problem (I'll check), the biggest step was solving for x.

[#1procrastinator]

The original problem was for f(x) = 2e^x - 4, find the inverse function and then find the coordinates of the intersection of the two graphs