Domain & Range Problem: Solve for a

Given that the domain and range of the function f(x) = sqrt{ax^2 + x} are the same set, we solve to find the value of a.
Here, we take the domain to mean the set of possible real inputs, x, so that sqrt{ax^2 + x} is a well-defined real number.
00:00 Intro
00:50 a = 0
01:37 Positive a
03:31 Negative a: Domain
04:42 Negative a: Range
06:53 Solving for a

[@Bayerwaldler]

Surprisingly interesting. I like the flawless presentation.

[@omograbi]

Why don't we try to equate the domain of both the function and it's inverse, i think the solution would be the same but from another perspective.

[@MyOneFiftiethOfADollar]

We? Why don't you do the honors of finding the inverse function to confirm your claim?

[@cosmicvoidtree]

A slight complication with that would be that sqrt(ax^2+x) isn’t injective so it doesn’t have a singular inverse function. You can work around this but it’s somewhat more difficult. It’s not a bad idea though.

[@PubicGore]

Interesting problem.

[@hvok99]

Very interesting problem, haven't considered such a thing before.

[@hvok99]

Just some extensions to this problems: which functions permit such solutions? are there functions which permit exactly 3 solutions? or arbitrarily many exact solutions? interesting stuff - excited to think about.

[@vvop]

Such a pretty problem, well done, thank you. It appears to be quite warm over there! Keep cool.😀

[@fabian4229]

Awesome video, thank you! 😊

[@kappascopezz5122]

If a=0, both the domain and range will be the non-negatives. If a>0, the parabola will be open upwards, so there will be some negative x in the domain which doesn't fit the range. For the rest of the calculation, assume a=0 x(ax+1) >= 0 With this, you get either x=0, and either ax+1=0. However, if the domain includes x

[@shrimp667]

Actually f(-1/(2a)) is sqrt(-1/(4a)) and you also made a mistake at the last part If 1=-a/2 then a=-2

[@kappascopezz5122]

@@shrimp667 you're right, I edited the comment to remove those mistakes

[@Umbra451]

Fascinating question! The mind of course wonders if we can extend this question to having x in C.

[@Ann-mj4xn]

No, the range would always be equivalent to the domain = C

[@LatinosErastis25]

I don't understand why we discarded a>0 as a possibility. Since if a > 0, we can just limit x to the domain [0, +inf). Upon which (ax^2 + x) is greater than or equal to zero, f(0)=0 and f(x) is strictly increasing with a range of [0, +inf).

[@LatinosErastis25]

We care about the constraint of Domain=Range being true. We don't have to restrict ourselves to only the cases where the Range is equal to the whole Domain. It works both ways, we can also restrict the Domain to the intervals at which it is equal to the Range. Maybe I am just answering a different question, I don't know, now I got all confused.

[@cameroncurtis7261]

Yeah the question does assume we "maximize" domain and range given "a", so any domain given a > 0 will include the interval (/infty , -1/a], which cant equal the range of the function. If we could limit the domain of the function, then yeah your point stands.

[@LatinosErastis25]

@@cameroncurtis7261I concluded that it basically comes down to semantics. By limiting the domain I am essentially making up a different function altogether, which is in no way what the problem asked for.

[@carterbagvitch5964]

I know this is not the answer, but if you let a=x the domain=the range. Just a lil cheeky answer

[@matze9713]

Solve for a what?

[@Bayerwaldler]

Maybe for a fistful of dollars ? 🤔

[@James2210]

a=-7/2

[@sran2007]

Why can’t range be negative? Sqrt can be both positive or negative.

[@ultimatedude5686]

Generally when we use the square root as a function we are referring to the principal square root, which cannot be a negative number.

[@MyOneFiftiethOfADollar]

Perhaps the most common conceptual mistake in the teaching of mathematics. Sqrt(x) would not be a function if it were both positive and negative. Consider sqrt(9) In your universe sqrt(9) = 3 and sqrt(9) = - 3 Like many, you are confusing this with the solutions to x^2=9 which are 3=sqrt(9) and -3=-sqrt(9) Hope this helps.

[@RexxSchneider]

@@MyOneFiftiethOfADollar And that's perhaps the most common mistake overall in mathematics: thinking that a convention is a fact. See en.wikipedia.org/wiki/Multivalued_function

[@MyOneFiftiethOfADollar]

@@RexxSchneider if you want to believe sqrt(9)=3 and sqrt(9)=-3, then you don’t understand the notion of a well defined function properly. Your usage of the term convention is dubious in this case. Surely you understand the difference between the solutions of x^2 = 9{-3,3} and sqrt(9)=3 ?

[@RexxSchneider]

@@MyOneFiftiethOfADollar As usual, I end up explaining to spoon-fed kids that the square root of 9 is just as much -3 as it is 3. There is nothing in maths that forces us to make functions single-valued. Of course, there are numerous applications where it is _convenient_ to have a function return only one value. But there is nothing inherently wrong with the concept of a function returning a set of values. You are surely capable of reading, so I assume you were simply dismissive of anything that doesn't fit what you've been taught, otherwise you might have read the Wikipedia article I pointed you to. If you are working in a field where it is advantageous to treat functions as singled-valued -- real analysis is a good example -- you still have to accept that you often have to decide which of a possible set of values you are going to take as the principal (or sole) value of the function. Making such a choice, for example taking the non-negative square root of a real number, is what is called a "convention" in English. So now you have a concrete example of what the word means. Hope that helps.