How to find derivative critical points that DON'T use ref. angles?

In summary: You then need to find the solutions to this equation, which is a standard exercise in algebra and trigonometry.
  • #1
Ascendant78
328
0

Homework Statement


I am trying to figure out how to solve for problems that give a function and ask for the critical points. However, the ones I am having difficulty with are the ones that have multiple trigonometric functions included in them.

Homework Equations


Find the critical points of the following function on the given interval:
f(x) = 8sin(x)+3cos(x) ; [-2∏, 2∏]

The Attempt at a Solution


Once I find the derivative, I grouped the trigonometric functions onto one side to make:
8/3 = tan(x)

However, I am at a loss as to how I can solve for x here? Our instructor never explained how we can solve for trigonometric functions without using the reference angles. I am trying to brush up on this stuff since I have calc II this fall, but am finding a couple things like this that we didn't cover.

Please keep in mind when providing your solution and method to that solution that I have only taken calc I up to this point. Just throwing that out there in hopes that you take this into consideration and don't throw things out there that might go over my head at this point of my education. Thank you.
 
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  • #2
##tan^{-1}(tan(x)) = x##

Use the inverse tangent ( arc tangent if you prefer denoted arctan(x) ).

So also ##arctan(tan(x)) = x##
 
  • #3
Draw a rough tan graph over the interval you are working with, notice where tan(x) > 0. Find a general formula for x using the ##tan^{-1}## function (tan is periodic with period π).

The most important thing to remember with these is to be absolute certain about the range of the inverse function.
 
  • #4
You should be aware that the statement of Zendrina is highly misleading and very dangerous! The usual real-valued arctan function is defined as
[tex]\arctan:\mathbb{R} \rightarrow (-\pi/2,\pi/2),[/tex]
which is known as the main branch of the arctan function in complex function theory.

So for a given number [itex]y \in \mathbb{R}[/itex] you find one specific angle [itex]x \in (-\pi/2,\pi/2)[/itex] such that [itex]\tan x=y[/itex]. Now, you want to find not only this one specific solution for you equation but all. It's however easy to find all these (infinitely many!) other solutions by thinking carefully about the properties of the tangens. It's best to draw the function [itex]y=\tan x[/itex] and also the function [itex]y=8/3[/itex] to see what's going on here!
 
  • #5
Thank you for the feedback. As far as the graphs of each and what is happening, that part is easy to me. However, the problem wants the exact values, which is where I am having a hard time with this. Is it something you would typically need a graphing program or calculator for, or is there a way to do it without one? Either way, I'm thinking now might be a good time for me to learn how to find graph intersections on my calc.
 
  • #6
Ascendant78 said:

Homework Statement


I am trying to figure out how to solve for problems that give a function and ask for the critical points. However, the ones I am having difficulty with are the ones that have multiple trigonometric functions included in them.


Homework Equations


Find the critical points of the following function on the given interval:
f(x) = 8sin(x)+3cos(x) ; [-2/∏, 2/∏]


The Attempt at a Solution


Once I find the derivative, I grouped the trigonometric functions onto one side to make:
8/3 = tan(x)

However, I am at a loss as to how I can solve for x here? Our instructor never explained how we can solve for trigonometric functions without using the reference angles. I am trying to brush up on this stuff since I have calc II this fall, but am finding a couple things like this that we didn't cover.

Please keep in mind when providing your solution and method to that solution that I have only taken calc I up to this point. Just throwing that out there in hopes that you take this into consideration and don't throw things out there that might go over my head at this point of my education. Thank you.

You wrote the interval as [-2/∏, 2/∏]. Do you mean [-π/2,π/2] or [-2π,2π]?
 
  • #8
Ascendant78 said:

Homework Statement


I am trying to figure out how to solve for problems that give a function and ask for the critical points. However, the ones I am having difficulty with are the ones that have multiple trigonometric functions included in them.


Homework Equations


Find the critical points of the following function on the given interval:
f(x) = 8sin(x)+3cos(x) ; [-2∏, 2∏]


The Attempt at a Solution


Once I find the derivative, I grouped the trigonometric functions onto one side to make:
8/3 = tan(x)

However, I am at a loss as to how I can solve for x here? Our instructor never explained how we can solve for trigonometric functions without using the reference angles. I am trying to brush up on this stuff since I have calc II this fall, but am finding a couple things like this that we didn't cover.

Please keep in mind when providing your solution and method to that solution that I have only taken calc I up to this point. Just throwing that out there in hopes that you take this into consideration and don't throw things out there that might go over my head at this point of my education. Thank you.

You can always reduce ##8\sin(x)+3\cos(x)## to ##a \sin(x+b)## for constants ##a,b## that are related to 8 and 3; this is a very standard trick/method, which you would be well advised to learn. Anyway, that reduces your problem to one involving a single trig. function.
 

Related to How to find derivative critical points that DON'T use ref. angles?

1. What are derivative critical points?

Derivative critical points are points on a graph where the slope, or rate of change, of a function is equal to zero. These points are important because they can give us information about the behavior of the function, such as where it reaches a maximum or minimum value.

2. How do I find derivative critical points?

To find derivative critical points without using reference angles, you will need to use calculus. First, find the derivative of the function using the rules of differentiation. Next, set the derivative equal to zero and solve for the variable. The solutions to this equation will be the critical points.

3. Can I use reference angles to find derivative critical points?

Yes, reference angles can be used to find derivative critical points, but this method is typically used for trigonometric functions. For other types of functions, it is more common to use calculus to find the critical points.

4. Why is it important to find derivative critical points?

Finding derivative critical points can help us understand the behavior of a function, such as where it reaches a maximum or minimum value. This information can be useful in many areas of science and engineering, such as optimization problems and analyzing real-world data.

5. Are there any shortcuts or tricks for finding derivative critical points?

There are some techniques that can make finding derivative critical points easier, such as using the first or second derivative test. However, these techniques are not always reliable and it is important to understand the underlying concepts of finding critical points through calculus.

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