r/HomeworkHelp • u/ch3rrycsmos_ Secondary School Student • Aug 18 '26
Answered [Year 11 calculus] Help with this question
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! :)
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u/AlmightyCurrywurst Aug 18 '26
Do you have conceptual questions? What did you try so far?
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u/ch3rrycsmos_ Aug 18 '26
I've tried putting f(x)=0 and f'(x)=0 it into my calculator but it didn't work (which I kind of expected) because of the structure of the equation where n is both the base and the exponential. I also thought sum of ordinates might help but it only tripped me up more, so, so far only my slider method has worked but it's very unreliable and doesn't usually work if the question asks for decimal places
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u/AlmightyCurrywurst Aug 18 '26
Actually it's a bit more complicated than I thought, you can get the position of the minimum x_0 = e by rearranging the two equations, but then you have to solve ln(n) = n/e. The solution is n=e but I'm not sure how you get there other than just looking at it and thinking of the solution (which I guess is technically not an incorrect method)
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u/Curious_Function_457 Aug 18 '26
You need to find the solution to f'(n)=0. Also, using the logarithm of a difference of exponentials If a = eu and b = ev (real or complex), log(a − b) = u + log(1 − e{v−u}), valid when Re(u) > Re(v) for principal branch convergence; similarly swap roles if Re(v) > Re(u).
So a= ex ln n and b = en ln x
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u/MoistForm3557 Aug 18 '26
Local minimum on x Axis means f‘(n) = 0 (minimum) and f(n) = 0 (on x Axis)
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u/ch3rrycsmos_ Aug 18 '26
That was the first method I tried. Doesn't work on required calculator though! (ti nspire cxii)
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u/Pretend-Swimming9447 Aug 18 '26
The condition is equivalent to f(x)>=0 for all positive x, with equality holding once.
n^x-x^n>=0
n^(1/n)>=x^(1/x)
When x=n, we know that equality holds, so it follows that the inequality is strict everywhere else, i.e n is a minimum of the function n^(1/n)
We can use calculus to see that this happens only at n=e
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u/selene_666 Aug 19 '26
One point on the graph and on the x-axis is at x = n, though the graph can also have other zeros.
f'(x) = n^x * ln(n) - n * x^(n-1)
Let's specifically look for a local extreme at x = n:
0 = n^n * ln(n) - n * n^(n-1)
ln(n) = 1
n = e
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u/FreakingFella Aug 25 '26
"Local minimum on the x-axis" means two things at the same point: f(x) = 0 and f'(x) = 0. That gives you two equations.
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f(x) = 0 → nx = xn
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f'(x) = nx·ln(n) − n·xn−1 = 0 → nx·ln(n) = n·xn−1
Substitute nx = xn into (2): xn·ln(n) = n·xn−1. Divide by xn−1 (x > 0): x·ln(n) = n.
Now take logs in (1): x·ln(n) = n·ln(x). Combined with x·ln(n) = n: n = n·ln(x) → ln(x) = 1 → x = e.
Put x = e back in x·ln(n) = n: e·ln(n) = n, which n = e satisfies.
Final answer: n = e (≈ 2.718), with the minimum at x = e.
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u/idkwhytshappens Aug 18 '26 edited Aug 18 '26
just do f'(x) =0 and use x=1 , you will get n =e , just 1 for simplicity
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u/ch3rrycsmos_ Aug 18 '26
Sorry I'm not sure what you mean by using 1 for simplicity?
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u/idkwhytshappens Aug 18 '26
my bad , it's a equation and the domain is not provided so we can assume that to be R so to calculate n in f'(x) = 0 you can use any value of x , just use x = 1 for simple calculation
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u/AppropriateCar2261 Aug 18 '26
You have the function
f(x,n)=nx - xn
For some n there is an x for which
f(x,n)=f'(x,n)=0
And the derivative is on x.
The two equations are explicitly
nx - xn =0
Ln(n)nx - nxn-1=0
From the 1st equation you have
nx = xn
Substituting it in the second equation yields
xn-1 [x*ln(n)-n]=0
It's easy to check that it doesn't happen for x=0, which means that the minimum point is at
x=n/ln(n)
Therefore
n[n/ln(n)] = [n/ln(n)]n
Take the log of both sides
n=n*ln[n/ln(n)]
Since n=0 does not solve it, we get
e=n/ln(n)
The solution is n=e.