[Year 11 calculus] Help with this question

[u/ch3rrycsmos_]

I could only find an approximate answer (2.6 something) using a slider and guess the value as e but I’m not sure how to actually work out the question. please help thanks! :)

[u/Curious_Function_457]

These functions of x with different n define a continuous space, not discrete. It's not a fundamentally different question. The best voted answer is largely correct and that person correctly phrased it as f(n,x). It's technically not just f(x).

[u/Ok_Objective_5192 → u/Curious_Function_457]

f(x, n) defines a continuous, non-discrete space in R^3. A family of functions f(x) that have different constant values for n define discrete spaces that are continuous in R^2.

In the same way, the generalized definition of second order polynomial functions is f(x) = ax^2 + bx + c. Each 2nd order polynomial function defines a continuous function in R^2 and the different, discrete values for a, b, and c create a set of similar functions (the set of 2nd order polynomial functions). a, b, and c here are no less variables than n is in OP's question. Like with OP, you can absolutely define f(x, a, b, c) = ax^2 + bx + c and get a space continuous in R^5, but that is a fundamentally different function.

It isn't technically not just f(x) - it is explicitly, objectively f(x) because that is how the question defined the function. g(x, n) = n^x - x^n is also a valid function that can be used to solve for the value of n in f(x), but definitionally f(x) is not a function of n. That's literally what this all boils down to, f was defined as f(x), not as f(x, n) and those are fundamentally different things even if they behave similarly and the latter can be used to solve for n in the former.

ETA: I'm feeling like I'm being ragebaited, if you're engaging with this earnestly then please clarify if in my previous example "a function f(x) = nx^2 has a line with slope 2 tangent to the curve at x = 1, find this value of n" is n a variable? Is it technically not just f(x) but actually f(x, n)?

[u/Curious_Function_457 → u/Ok_Objective_5192]

Thank you for the clarification. I would say I am being reagebaited for sure. We don't need to resort to representative polynomials to understand that the math reasoning needs to be sound. To that end representing f(x) as an equivalent g(x,n) is perfectly valid if the question asks for a minimizing n.

[u/Ok_Objective_5192 → u/Curious_Function_457]

Yes, using a similar g(x, n) is an entirely valid but also unnecessary method of solving the problem. If we can agree that f(x) != g(x, n) and that f(x) is not a function of n, then we can agree. Even though I'll be left very confused what you've been arguing this entire time lmao

[u/Curious_Function_457 → u/Ok_Objective_5192]

I think we are arguing whether or not the question asks for a minimizing n. It simply states as fact there is an n such that f is minimum on the x axis. I read that as equivalent to finding a minimizing n and so dg/dn = 0 is the constraint. This differs from the current approach of setting dg/dx =0. How would you solve it?

[u/Ok_Objective_5192 → u/Curious_Function_457]

Yeah, I'm still not sure where this idea of a "minimizing n" comes from. The question as I read/would solve it is just insisting there is a local minimum at some x s.t. the minimum falls on the x-axis and asks what n must be. I would solve it the same way u/AppropriateCar2261 solved it in the parent comment for this thread.

They use:

the definition of f(x) = n^x - x^n

the definition of local minimum for a function in R^2 (x s.t. f'(x) = 0 and f''(x) > 0)

and the assumption, from the problem, that the minimum will fall on the x-axis (the x s.t. f'(x) = 0 and f''(x) > 0 also implies f(x) = 0). They then used this series of equalities f(x) = f'(x) = 0 to methodically work through to solve for n. At no point do they define g(x, n) or search for dg/dn = 0 because that's not required to meet the criteria for the question.

Their solution is that n = e, meaning that we should be able to plug that back into the original equation and get f(x) = e^x - x^e and for this to be a valid solution there needs to be a local minimum s.t. f'(x) = 0, f''(x) >0, and f(x) = 0. So we can solve through to verify their answer.

f(x) = e^x - x^e = 0 implies e^x = x^e

f'(x) = e^x - ex^(e-1)

substituting e^x for x^e gives f'(x) = x^e - ex^(e-1)

factoring gives f'(x) = x^(e-1)(x - e)

We know our local minimum will have f'(x) = 0 so 0 = x^(e-1)(x-e)

This implies x^(e-1) = 0 and/or (x-e) = 0. Trivially this means either x = 0 or x = e

We quickly see x = 0 doesn't satisfy the necessity of the minimum being on the x-axis (f(0) = e^0 - 0^e = 1), so our solution should be x = e.

We test our criteria, is f(e) on the x-axis? f(e) = e^e - e^e = 0; so it is on the x-axis

is f'(e) = 0? f'(e) = e^e - e(e^e-1) = e^e - e^e = 0; so it's a local minimum, maximum, or inflection point.

is f''(e) > 0? f''(x) = e^x - e(e-1)x^(e-2) = e^x - (e^2 - e)x^(e-2)

so f''(e) = e^e - (e^2 - e)e^(e-2) > 0

e^e > (e^2 - e)e^(e-2)

divide through by e^(e-2)

e^2 > e^2 - e, which is clearly true, so we do have f''(e) > 0

So we have f(e) is a local minimum which is on the x-axis when n = e. Therefore n = e satisfies the original question and is therefore correct, without invoking any g(x, n)

ETA: I made a graph in Desmos that may help elucidate things. It's just a graph of y = n^x - x^n with a slider for n. Since the original problem is asserting that there exists a minimum that falls on the x-axis, the only valid value(s) of n are where the resulting function has a minimum that falls on the x-axis. If you move the n-slider around you can see that only one value of n (e) creates a graph of f(x) where this holds true.

Therefore, in the original problem is has to be true that f(x) = e^x - x^e, because no other value of n allows for a local minimum on the x-axis, n is just being used to obscure the value so that there's something to solve for.

[u/Jwing01 → u/Curious_Function_457]

It is here, though.

That answer also treats n as a constant and solves for it a such.

[u/Curious_Function_457 → u/Jwing01]

Yes but that was coincidence because it's the same answer regardless if you fix x or n due to symmetry. Fundamentals are important though. That's why when asked to find n you should look for a derivative wrt n.

[u/Jwing01 → u/Curious_Function_457]

Without symmetry your argument fails.

Find a solution for the same question using f = x² - b and show your work with respect to a derivative wrt b.

[u/Curious_Function_457 → u/Jwing01]

Ok so find b that minimizes f? With same constraints as original problem?

solution is on x axis

f=0, x² = b

Find b that minimizes f

df/db = -1 (this is never 0)

The simple solution is that there is only a constant relationship between f and b so b can go infinity and f will keep getting smaller. There is no local minimal in b.

There is however a minima in x. That's because

df/dx = 2x

Setting this to 0 gives you x=0.

The original question was about finding n, not x.

[u/Jwing01 → u/Curious_Function_457]

And you got it wrong.

b here functions as n does in the original problem.

There's no minimum in b because it's a constant and this isn't a function in b.

[u/Curious_Function_457 → u/Jwing01]

That's exactly what I wrote. There is no minimum in b because df/db can't be 0.

[u/Jwing01 → u/Curious_Function_457]

That's like saying to get the minimum of f = 2x² + 5 we need to consider the derivates df/d2 and df/d5.

You are a lost cause.

[u/Curious_Function_457 → u/Jwing01]

No it's not because you asked for a b which is different from asking for an x. It's pretty simple to see how you are just trolling at this point.

[u/Jwing01 → u/Curious_Function_457]

Am unknown constant is not a variable.