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General method for determing max/min within a range? by Fucking Bellyburk - Thu, 28 Jun 2018 07:51:21 EST ID:rGLueGvy No.79150 Ignore Report Quick Reply
File: 1530186681836.jpg -(16201B / 15.82KB, 500x500) Thumbnail displayed, click image for full size. 16201
Does anyone please know of a general way to determine if there's any local max and min *within a set range of X-coordinates*?

As we can see, there's a local max somwhere around -1 and a local min somwhere around 3 and none whatsoever in the range 1-2.

But how do I prove that there's no local max nor min in that range?
Lydia Claystone - Thu, 28 Jun 2018 12:41:52 EST ID:ggkjpE8I No.79151 Ignore Report Quick Reply
u gotta plug in values of x over the range to detect a pattern. also i think if u take the derivative of the function then anywhere f'(x) = 0 is a max or a min but its been a while since i did calculus
press - Fri, 29 Jun 2018 08:34:10 EST ID:2alAeOJc No.79152 Ignore Report Quick Reply
is the range open or closed?
press - Fri, 29 Jun 2018 08:36:33 EST ID:2alAeOJc No.79153 Ignore Report Quick Reply
because if your range is closed, your function absolutely has a min / max on that range since its a continous function.
trypto - Sat, 30 Jun 2018 18:28:12 EST ID:OdR7meD+ No.79154 Ignore Report Quick Reply
If the interval is open, there's still a min/max, since a limit can be considered a min/max. For example, the global maximum of x^2 is infinity, not undefined.

OP, I don't think there's a 'general way' that works with all functions, because that's an extremely large category.

There's the standard algebra 2 method for vanilla functions. First, you check the intervals (beginning and end - this can be a finite range or from -inf to +inf.). Then you take the first derivative, set it to 0 and solve. Or maybe it's the second derivative. I forget. Google "points of inflection", and you can probably find the method me and Lydia are talking about.

That vanilla method won't work for a lot of functions. They need to be continuous and differentiable everywhere for that to work. So it's no good with fractals, step-functions, non-continuous functions (like the tangent in your OP), etc.
trypto - Sat, 30 Jun 2018 18:34:43 EST ID:OdR7meD+ No.79155 Ignore Report Quick Reply
Actually, I googled it for you.

Take the first derivative, set it equal to 0. Solve. That gives points of inflection.

Then you take the second derivative, set it equal to zero. Solve. That tells you if it's a maximum or a minimum.

For example, x^2. First, check the bounds. I think we all agree for -inf, the answer is inf. And for +inf, the answer is inf. That's your global maximum.

f(x) = x^2

f'(x) = 2x

0 = 2x
x = 0

That means there's a point of inflection at x =0. Now let's see if it's a global maximum or minimum (even though it's pretty obvious).

f''(x) = 2

Because that's positive, it means it's a minimum.

Hope that helps.
Oliver Grandstone - Mon, 02 Jul 2018 15:44:38 EST ID:rGLueGvy No.79158 Ignore Report Quick Reply
I'm looking for *local* max/min in trig-functions. I know that if the x-range is long enough, there will be *many* local max/min.

For instance this function has local max/min all over the place and repeats itself every 3600 x:

f(x) = sin x(sin (x/10))

I know there's a local max between x(1500) and x(1530). But how do I prove it by doing a f'(x) for only that range?
trypto - Tue, 03 Jul 2018 11:21:20 EST ID:OdR7meD+ No.79159 Ignore Report Quick Reply
I think it still works the same way. Take derivative. Set equal to 0. Solve within your range. That's a lot of trig functions to fuckin deal with. Here's wolfram alpha's solutions:


Solutions are at the bottom. Hit the 'exact solutions' to see what kinda algebra is needed for that function. So obviously the way to do this is numerically. Look at the original function

Then look at the approximate solutions in the last page. You can see that the local maximum for that range is at:
For which the function returns: 0.987810970335936684937

Or, in one easy step: https://www.wolframalpha.com/input/?i=maximum+sin(x)(sin(x%2F10)),+1500%3Cx%3C1530

Maybe that's cheating, but fuck trying to analytically solve that.
Simon Sannermutch - Wed, 04 Jul 2018 06:43:52 EST ID:rGLueGvy No.79160 Ignore Report Quick Reply
Thanks a lot! :-)
Cedric Gussleserk - Mon, 09 Jul 2018 04:18:54 EST ID:ggkjpE8I No.79164 Ignore Report Quick Reply
the fact that this thread about mathematics was posted on /chem/ is only further evidence of the genius of my idea of a /stem/ board

bress - Fri, 13 Jul 2018 13:01:59 EST ID:DbJ7MA1o No.79171 Ignore Report Quick Reply
1531501319502.jpg -(129931B / 126.89KB, 800x449) Thumbnail displayed, click image for full size.
the fact that youve posted that is further evidence of the genius of the idea to establish a /shill board. naw just kidding.

concerning the analytic component of the question, im kinda ashamed it took me this long to think of, you could take the derivative via product rule (not sure if another approach would deliver something more moldable) and then reason examine the zero points of the derivative and second derivative.

im a bit confused as to what the actual function is, should it - as mentioned before - be y= sin(x) sin(1/10), the derivative itself will be a sum of two periodic products, meaning youd have to evaluate cases where either both terms are zero or one term is the negative of the other. depending on the range the answer should be given in multiples of some fraction of pi (kirt, tex support when)

during my analytics classes giving a numeric answer was a telltale sign of misunderstanding the task

And should your task be to prove that the function has a local min/max on a given closed range you could simply show that the function is continous (as a composition of continous functions) and then a local min/max has to exist.

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