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Anonymous 18/06/28(Thu)13:01 No. 16668 ID: c6e9b2

File 153018367186.jpg - (15.82KB , 500x500 , 3bfc065f.jpg )

Yay /sci/!

Does anyone please know of a 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?

Anonymous 18/07/01(Sun)09:09 No. 16670 ID: af0184

First, what class are you taking? What level mathematics are you looking at?
Then, what is the f(x)?

Hopefully, calculus I. Take the derivative; set f(x)'=0, solve for all x values. That proves the only time slope=0 can be at those specific points (watch out for those pesky imaginary numbers).

We could also do a differential equation, i think. It looks like we have enough information from the graph to knock one out. However, its been awhile since I did one. Id need to review some notes.

Anonymous 18/07/25(Wed)08:59 No. 16682 ID: c6e9b2

Not taking any class and don't know my level.

And what if it's f(x)=(sin 10X)+X ?

That will be a constantly increasing sin-wave with no global maximum and a lot of local maximums.

Anonymous 18/07/25(Wed)21:33 No. 16684 ID: be6f8f

Given a closed interval of the domain, any continuous function *necessarily* has both a global maximum and a global minimum[1], since the closed edges of the interval are candidates to being global maximum and minimum. For example f(x) = x^2 in the interval [-1; 2] has a global minimum at x = 0 and a global maximum at x = 2. However, it has no global maximum in the intervals [-1; 2) and (-1; 2). On the other hand, f(x) = 1/x has no maximum or minimum in any of those intervals.

Discontinuous functions may not have any extrema even in closed intervals. For example, f(x) = x/x + x, interval [-1; 0].

There's no general algorithm to prove that a given function has no extrema in a given range. As >>16670 mentioned, the zeroes of the derivative is a good place to start, but it doesn't end there. Some functions have extrema and are continuous but nowhere differentiable[2]. Some have extrema and are nowhere continuous[3].

As for your example of f(x) = sin(10 * x) + x:

acos: [0, pi] -> Reals
acos(y) = the value of x such that cos(x) = y

f'(x) = 0
10 * cos(10 * x) + 1 = 0
10 * cos(10 * x) = -1
cos(10 * x) = -1/10
x in {(pi * (k * 2 - 1) - acos(1/10)) / 10 | k in Integers}

Now you have the zeroes of the derivative. To show that those points are local extrema, you just need to show that the derivative at those points passes from positive to negative or viceversa (e.g. x^2 has a zero a x = 0, but it doesn't change sign. x^3 does). I'll leave this as an exercise, but it can be done using properties of cos().

[1] https://en.wikipedia.org/wiki/Extreme_value_theorem
[2] https://en.wikipedia.org/wiki/Weierstrass_function
[3] Let f(x) = (if x is rational then x else -x), then f(x) has a maximum but no minimum in the interval [0; 1] (f(1) = 1 and it takes no larger value in that interval).

Further reading:

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