brightsky's Noob-tastic Question Thread

[Linkage1992]

yeah i think the answers wrong, because that's what I got too...

[brightsky]

It's quite a misleading question, but defying all initial intuition, Jess is taking without replacement (as kamil suggested in the other post). Taking that into account, you have the required answer.

[brightsky]

I was just like to know, is there an efficient way to approach questions that ask you to find the intersection between two circles? It seems really draining and error-prone to use the substitution method. :/

[brightsky]

Find the range of . My approach was:







We want , which yields .

This method seems a bit dodgy. Just like to know, is this usually how one would approach these questions, or is there a better more generic way of doing it?

Thanks.

[dooodyo]

Hey Brightsky,

How did you get -1 < k < (1/6) as the range?

I thought that you could just resolve the equation into partial
fractions and then simply just find the range from the new
equation?

[brightsky]

Hey Brightsky,

How did you get -1 < k < (1/6) as the range?

I thought that you could just resolve the equation into partial
fractions and then simply just find the range ?

Nah, I don't think you can simplify it into partial fractions. I got that by multiplying both sides by (x^2 + 6) and then solving the resulting quadratic to find the restrictions for y. But seriously, it's not an obvious method, and it just seems very obscure. I'm sure there is a better way to do these types of questions (a general idea that can be applied to all 'range' finding).

[kamil9876]

I think that's a good solution. There is no real way for ALL functions but I will show you an approach that works for differentiable functions where the domain is a CLOSED interval . Just find the minimum and maximum using differentiation (don't forget to compare them with the endpoints) and then the range is simply [m,M] since your function is continous (google intermediate value theorem and extreme value theorem if you want). Now this approach doesn't generalise nicely to an open interval or to the whole or (like in your case) or even worse it is crap for something like . But what you can do if the domain is whole of is the following if your lucky: if the limits as x goes to infinity or -infinity are in (m is the smallest local minimum, M is the largest local maximum that you get from differentiating(if they exist)) then you can conclude that the range is indeed .

This should work for your example as the limits are 0 and should be inside the [m,M] that you get from differentiating (note that is always a subset of the range if the function is differentiable over an interval (no gaps like 1/x which has  gap at x=0))

[brightsky]

How to find the domain and range of:



More specifically in regards to the domain,

I got and and solve simultaneously, but if you take the up into the denominator, i.e.

, letting gives you different solutions...am I missing something?

[pi]

I suck at these questions, but:







I think for domains/ranges, you always should use the original function and not rearrange it


EDIT: stupid error fixed

[xZero]

I suck at these questions, but:






I think for domains/ranges, you always should use the original function and not rearrange it

shouldnt it be

if we combine it,

you'll get
and

now if both brackets are negative, you can still get but if we consider the denominator, , hence both brackets must be positve. So we change it to , which is the same as the answer we get before we combine it into a fraction

[brightsky]

Thanks for that, yeah that's what I got, but I was a bit skeptical about giving two solutions. I think I'm getting confused over something trivial/obvious, but my brain isn't working...:/

I think for domains/ranges, you always should use the original function and not rearrange it

What if the original question was to find the domain/range of ?

Also any help on the range?

[brightsky]

I suck at these questions, but:






I think for domains/ranges, you always should use the original function and not rearrange it

shouldnt it be

if we combine it,

you'll get
and

now if both brackets are negative, you can still get but if we consider the denominator, , hence both brackets must be positve. So we change it to , which is the same as the answer we get before we combine it into a fraction

The quadratic inequation gives and .

[xZero]

really?


and

and

now we can consider when they are both negative since 2 negatives will become positive, hence greater than 0. however (2x-1) must be greater than 0 so in this case we can only consider when both brackets are positve, hence the answer

[pi]

really?


and

and

I thought you were not able to solve quadratic inequations like that. I though a mental sketch graph was necessary... Which would make and correct.

EDIT: Wolfram Alpha agrees too

[brightsky]

really?


and

and

now we can consider when they are both negative since 2 negatives will become positive, hence greater than 0. however (2x-1) must be greater than 0 so in this case we can only consider when both brackets are positve, hence the answer

Yeah, but we must consider the double negative case as well. The case you just gave is basically just one solution, i.e.g . The other one is redundant.