Using Rolle's Theorem to Find Extremes of f(x) = sin 2x

[Jimmy84]

Homework Statement

I need to use Rolle's theorem to find analitically the maxium and the minimun extremes on the interval [0 , 1/2pi] .


So I think I must first find the derivative of the fuction and to solve for x


the function given is f(x) = sin 2x

Homework Equations

The Attempt at a Solution

Im not sure about how to start the product rule

is it like this?

sin x (2x)

then the product rule would be

sin x (2) + 2x (cos x) ?



I have no idea about how to solve for x.

 

[CompuChip]

There is no product rule to be applied.
sin(2x) is something different than sin(x) * 2x.
If you call f(x) = sin(x), this is easier to see (f(2x) and 2 x f(x) are completely different things, and the derivative of f(2x) is not 2f(x) + 2x f'(x), or something like that).

What you are looking for is the chain rule.

 

[Jimmy84]

There is no product rule to be applied.
sin(2x) is something different than sin(x) * 2x.
If you call f(x) = sin(x), this is easier to see (f(2x) and 2 x f(x) are completely different things, and the derivative of f(2x) is not 2f(x) + 2x f'(x), or something like that).

What you are looking for is the chain rule.

ok using f ' (g) g'(x) resulted on cos(2x) 2 that is I think 2 cos 2x .



Now I am not sure about what to do next in order to look for the maximum and minimum extremes on the function.

On previous examples that were not trigonometric functions I solved for x on the original function and I also solved for x on the derivative and I found the critical points that way.

How can I find the critical points of f ?

 

[Mark44]

What does Rolle's Theorem say? You have a function, f(x) = sin(2x) for which f(0) = f(pi/2) = 0, and f is a differentiable function (you found its derivative).

 

[Jimmy84]

What does Rolle's Theorem say? You have a function, f(x) = sin(2x) for which f(0) = f(pi/2) = 0, and f is a differentiable function (you found its derivative).

So the zeros are 0 and pi/2 . The function is continuous on [0,pi/2] and differentiable on
(0,pi/2)

Well I graphed the function and the maximum extreme is obiously 1 but if I evaluate
2cos(2x) that is the derivtive of f by 0 I get

2cos(0) = 2

what am I doing wrong?

 

[Mark44]

So the zeros are 0 and pi/2 . The function is continuous on [0,pi/2] and differentiable on
(0,pi/2)

What does Rolle's Theorem conclude?

Well I graphed the function and the maximum extreme is obiously 1 but if I evaluate
2cos(2x) that is the derivtive of f by 0 I get

2cos(0) = 2

what am I doing wrong?

Why would you want to evaluate f'(0)?

 

[ideasrule]

You've shown that the derivative of sin(2x) is 2 at x=0. Nothing in math says that the derivative can't be larger than the maximum value of the function.

Draw out a rough sketch of sin(2x). What's the slope of the tangent line equal to at the highest and lowest points?

Note that you should be able to solve this question intuitively, without using calculus. What's the maximum and minimum values of any sine function?

 

[Jimmy84]

You've shown that the derivative of sin(2x) is 2 at x=0. Nothing in math says that the derivative can't be larger than the maximum value of the function.

Draw out a rough sketch of sin(2x). What's the slope of the tangent line equal to at the highest and lowest points?

Note that you should be able to solve this question intuitively, without using calculus. What's the maximum and minimum values of any sine function?

The derivative of any maximum or minimum equals to 0
and the pourpose of Rolle's theorem as far as I know is to show that there is a number c in the interval [0,pi/2] such that f '(c) = 0

Im being asked to find c as well. how can I find c ?

I think c is called a critical point as well.

 

[Jimmy84]

by the way the answer in my book is 1/4pi I don't understand how that happened.

 

[Mark44]

The derivative of any maximum or minimum equals to 0

This is not necessarily true. For example, the minimum value of y = f(x) = |x| is 0, and it occurs at x = 0, but f'(x) is never equal to 0 for this function.

and the pourpose of Rolle's theorem as far as I know is to show that there is a number c in the interval [0,pi/2] such that f '(c) = 0

Close. Rolle's Thm says that there is a number c in the open interval (0, pi/2) such that f'(c) = 0.

That means that for some number 0 < c < pi/2, 2cos(2c) = 0. Can you solve this equation?

Im being asked to find c as well. how can I find c ?

I think c is called a critical point as well.

 

[Jimmy84]

This is not necessarily true. For example, the minimum value of y = f(x) = |x| is 0, and it occurs at x = 0, but f'(x) is never equal to 0 for this function.

Close. Rolle's Thm says that there is a number c in the open interval (0, pi/2) such that f'(c) = 0.

That means that for some number 0 < c < pi/2, 2cos(2c) = 0. Can you solve this equation?

Thanks, that's right I think that f '(x) = |x| is undefined at x= 0 .

no, I can't solve that equation this is kind of new to me. that's the main issue I am having with this problem so I don't have a clue. I don't remember ever solving variable equations like that.

I guess that by guessing it could be factorized like this

2 (cos (c)) = 0

but that's about it.

 

[Jimmy84]

I don't think that this is right but anyway

2cos(2c) = 0

2c = 1/2 cos x = c = 1/4 cos x

but I don't understand where did pi came from in c = 1/4pi which is the original result of the book.

Can you give me a hint?

 

[Mark44]

Thanks, that's right I think that f(x) = |x| is undefined at x= 0 .

No, f(x) = |x| is defined for all values of x, but for this function, f' is not defined at x = 0.

no, I can't solve that equation this is kind of new to me. that's the main issue I am having with this problem so I don't have a clue. I don't remember ever solving variable equations like that.

I guess that by guessing it could be factorized like this

2 (cos (c)) = 0

but that's about it.

 

[Mark44]

I don't think that this is right but anyway

2cos(2c) = 0

2c = 1/2 cos x = c = 1/4 cos x

but I don't understand where did pi came from in c = 1/4pi which is the original result of the book.

Can you give me a hint?

Yes - it's not right.

2cos(2c) = 0
==> cos(2c) = 0

Now, I'm hopeful that you have seen the graph of y = cos(x).
For what values of x is cos(x) = 0? One of them is pi/2 and there are lots and lots of values.

The same values of x for which cos(x) = 0 are the same values of 2c, so what are the values of c?

 

[Jimmy84]

Yes - it's not right.

2cos(2c) = 0
==> cos(2c) = 0

Now, I'm hopeful that you have seen the graph of y = cos(x).
For what values of x is cos(x) = 0? One of them is pi/2 and there are lots and lots of values.

The same values of x for which cos(x) = 0 are the same values of 2c, so what are the values of c?

Ok, thanks for your time Mark, take care.