Minima of sin function with domain (a, b)

ndesk1900
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Homework Statement


Consider the function f(x) = sin (x) in the interval x ∈ (p/4, 7p/4). The number and location(s) of the local minima of this function are?

Homework Equations




The Attempt at a Solution


I have managed to get one of the minimas (i.e. 3p/2), however i am unsure about the the minima. If x ∈ [p/4, 7p/4], then it would be p/4 however since x ∈ (p/4, 7p/4) and p/4 is not part of the domain not sure what the minima would be. Here is an image to denote the same idea.

*Please note p is pie here.
 
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ndesk1900 said:

Homework Statement


Consider the function f(x) = sin (x) in the interval x ∈ (p/4, 7p/4). The number and location(s) of the local minima of this function are?

Homework Equations




The Attempt at a Solution


I have managed to get one of the minimas (i.e. 3p/2), however i am unsure about the the minima. If x ∈ [p/4, 7p/4], then it would be p/4 however since x ∈ (p/4, 7p/4) and p/4 is not part of the domain not sure what the minima would be. Here is an image to denote the same idea.

*Please note p is pie here.

This character - π - is pi, not pie.

For the function on its restricted domain, there is one maximum (plural is maxima) and one minimum (plural is minima). The usual techniques of calculus can be used to find these critical points.
 
But it is sufficient to know that sin(x) is always between -1 and 1 and is equal to -1 if and only if x= (4n+3)\pi/2where where n is any integer.
 
Mark44 said:
This character - π - is pi, not pie.

For the function on its restricted domain, there is one maximum (plural is maxima) and one minimum (plural is minima). The usual techniques of calculus can be used to find these critical points.

However, the question asked for *local* maxima/minama, so the endpoints could count as well (at least in standard optimization terminology, where a local min in an interval means that we do not look outside the interval, so can have a positive derivative if the local min is a left endpoint or a negative derivative at a right-hand end point). However, not all treatments agree with that terminology; it is more commonly used in optimization textbooks than in calculus textbooks, for example (because in the 'real world' we very often encounter such solutions to problems, and we need a way to talk about them in a common language). The OP will need to use terminology relevant to the course.

RGV
 
Ray Vickson said:
However, the question asked for *local* maxima/minama, so the endpoints could count as well
Understood.
However, the OP provided a graph from which it was obvious to the most casual observer that endpoints of the restricted domain were not going to contribute maxima or minima. For that reason, I didn't say anything about checking the endpoints, something that I normally do.
Ray Vickson said:
(at least in standard optimization terminology, where a local min in an interval means that we do not look outside the interval, so can have a positive derivative if the local min is a left endpoint or a negative derivative at a right-hand end point). However, not all treatments agree with that terminology; it is more commonly used in optimization textbooks than in calculus textbooks, for example (because in the 'real world' we very often encounter such solutions to problems, and we need a way to talk about them in a common language). The OP will need to use terminology relevant to the course.

RGV
 
Mark44 said:
Understood.
However, the OP provided a graph from which it was obvious to the most casual observer that endpoints of the restricted domain were not going to contribute maxima or minima. For that reason, I didn't say anything about checking the endpoints, something that I normally do.

The only reason they are not local min/max is that in this case the domain is an open interval instead of a closed one (and I assume the OP *really* meant what he wrote). Had the interval been closed, the endpoints would be local optima (but, assuredly, not global optima), inasmuch as they would respect the definition that x_0 \in [a,b] is a local minimizer (maximizer) of f(x) on [a,b] if f(x_0) \leq f(x) \;(\geq f(x)) \: \forall x \in [a,b] \text{ in a neighbourhood of } x_0.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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