Minima of sin function with domain (a, b)

In summary: 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
  • #1
ndesk1900
4
0

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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  • #2
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.
 
  • #3
But it is sufficient to know that sin(x) is always between -1 and 1 and is equal to -1 if and only if [itex]x= (4n+3)\pi/2[itex] where where n is any integer.
 
  • #4
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
 
  • #5
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
 
  • #6
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 [itex] x_0 \in [a,b][/itex] is a local minimizer (maximizer) of [itex]f(x)[/itex] on [itex][a,b][/itex] if [itex] f(x_0) \leq f(x) \;(\geq f(x)) \: \forall x \in [a,b] \text{ in a neighbourhood of } x_0.[/itex]

RGV
 

1. What is the definition of a "minima" in a sin function?

A minima is a point on a curve where the function reaches its lowest value, or the lowest point on a downward slope.

2. How do you find the minima of a sin function with a given domain (a, b)?

To find the minima of a sin function with a given domain (a, b), you can take the derivative of the function and set it equal to 0. Then, solve for x to find the critical points. Plug these critical points into the original function to find the corresponding y-values. The lowest y-value will be the minima.

3. Can a sin function have multiple minima within a given domain?

Yes, a sin function can have multiple minima within a given domain. This occurs when the function repeats itself and has multiple downward slopes.

4. How do the values of a and b affect the minima of a sin function?

The values of a and b determine the domain of the function, which in turn affects the number and location of minima. A smaller domain (a, b) will result in fewer minima, while a larger domain will result in more minima.

5. Are there any special properties of the minima in a sin function?

Yes, the minima of a sin function have some special properties. They occur at intervals of π and have a value of -1, which is the lowest value the function can reach. Additionally, the minima can be used to determine the amplitude and period of the function.

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