Prove that this function is nonnegative

[sergey_le]

Homework Statement

Prove that $x^2$/2>xcosx-sinx for all x≠0

Relevant Equations

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What I wanted to do was set f(x)=$x^2$/2 - xcosx+sinx And show that f(x)>0.
f'(x)=x(1+sinx)
First I wanted to prove that f(x)<0 in the interval (0,∞)
0≤1+sinx≤2
And thus for all x> 0 f'(x)≥0 and therefore f(x)≥f(0)=0
And it doesn't help me much because I need to f(x)>0

 

[Mark44]

Homework Statement:: Prove that $x^2$/2>xcosx-sinx for all x≠0
Homework Equations:: -

What I wanted to do was set f(x)=$x^2$/2 - xcosx+sinx And show that f(x)>0.
f'(x)=x(1+sinx)
First I wanted to prove that f(x)<0 in the interval (0,∞)

No, you need to prove that f(x) > 0 in that interval, just as you say above.

0≤1+sinx≤2
And thus for all x> 0 f'(x)≥0 and therefore f(x)≥f(0)=0
And it doesn't help me much because I need to f(x)>0

You showed that $f'(x) \ge 0$ for $x \in [0, \infty)$, which means that the graph of f is increasing on this interval. For the other part of this problem, show that $f'(x) \le 0$ for $x \in (-\infty, 0]$. Do you see how this helps you with the other part of the problem?

 

[sergey_le]

No, you need to prove that f(x) > 0 in that interval, just as you say above.
You showed that $f'(x) \ge 0$ for $x \in [0, \infty)$, which means that the graph of f is increasing on this interval. For the other part of this problem, show that $f'(x) \le 0$ for $x \in (-\infty, 0]$. Do you see how this helps you with the other part of the problem?

No
That's what I intended to do.
But my problem is with the one that can be x∈(0,δ) (when are δ>0) so that f'(x)=0 and then f(0)=f(x)=0 for ∀x∈(0,δ) And then what I have to prove is wrong.
Do you understand my problem?

 

[PeroK]

No
That's what I intended to do.
But my problem is with the one that can be x∈(0,δ) (when are δ>0) so that f'(x)=0 and then f(0)=f(x)=0 for ∀x∈(0,δ) And then what I have to prove is wrong.
Do you understand my problem?

Try to prove:

If $f(0)=0$ and $f'(x) > 0$ for all $x > 0$ then $f(x) > 0$ for all $x >0$. Even if $f'(0)=0$.

 

[FactChecker]

Do you know the Fundamental Theorem of Calculus? That is a good way to use properties of the derivative to prove things about the function.

 

[sergey_le]

Try to prove:

If $f(0)=0$ and $f'(x) > 0$ for all $x > 0$ then $f(x) > 0$ for all $x >0$. Even if $f'(0)=0$.

If that was the case then I had no problem .
But because it is ≤ Then there is the possibility that f'(x)=0.

 

[sergey_le]

Do you know the Fundamental Theorem of Calculus? That is a good way to use properties of the derivative to prove things about the function.

Yes but I can't use it in this course.
In my course it is Calculus 2

 

[PeroK]

If that was the case then I had no problem .
But because it is ≤ Then there is the possibility that f'(x)=0.

Why did you choose $\le$?

What I mean is this: suppose you wanted to show that $z > 0$, say. First you show that $z \ge 0$ and then say you are stuck. So, you need to go back and try to prove that $z > 0$.

 

[FactChecker]

Yes but I can't use it in this course.
In my course it is Calculus 2

Are you allowed to use the Mean Value Theorem?

 

[sergey_le]

Why did you choose $\le$?

What I mean is this: suppose you wanted to show that $z > 0$, say. First you show that $z \ge 0$ and then say you are stuck. So, you need to go back and try to prove that $z > 0$.

That's exactly my problem. Don't see a reason that z> 0

 

[sergey_le]

Are you allowed to use the Mean Value Theorem?

yes

 

[PeroK]

That's exactly my problem. Don't see a reason that z> 0

Okay, let's start at the beginning. You have defined the function:

$f(x) = \frac 1 2 x^2 -x\cos x + \sin x$

You want to show that $\forall x \ne 0: \ f(x) > 0$ Let's start with $x > 0$

Note that $f(0) = 0$. First, you have to differentiate $f(x)$. Can you do that?

 

[FactChecker]

I think that you can use the Mean Value Theorem to prove it. Remember that the slope of a tangent line is the derivative. Use a proof by contradiction. Assume that there is a point, a, where f(a)<0 and show that the assumption must be wrong.

 

[PeroK]

I think that you can use the Mean Value Theorem to prove it. Remember that the slope of a tangent line is the derivative. Use a proof by contradiction. Assume that there is a point, a, where f(a)<0 and show that the assumption must be wrong.

That almost works, except it must be shown that $f(x) > 0$.

 

[sergey_le]

I think that you can use the Mean Value Theorem to prove it. Remember that the slope of a tangent line is the derivative. Use a proof by contradiction. Assume that there is a point, a, where f(a)<0 and show that the assumption must be wrong.

If I use contradiction, So I have to assume that f(a)≤0 , And I don't see a reason why that f(a)=0

 

[sergey_le]

Okay, let's start at the beginning. You have defined the function:

$f(x) = \frac 1 2 x^2 -x\cos x + \sin x$

You want to show that $\forall x \ne 0: \ f(x) > 0$ Let's start with $x > 0$

Note that $f(0) = 0$. First, you have to differentiate $f(x)$. Can you do that?

I can show that f(x)≥0 .

 

[PeroK]

I can show that f(x)≥0 .

I know that. And you're asked to show that $f(x) > 0$.

Why don't you differentiate the function? We need to analyse the derivative.

 

[sergey_le]

I know that. And you're asked to show that $f(x) > 0$.

Why don't you differentiate the function? We need to analyse the derivative.

Also the derivative f'(x)≥0
I don't understand what you want me to do?
How do I show that f'(x)>0 And no ≥

 

[Mark44]

First, you have to differentiate f(x)f(x)f(x). Can you do that?

Why don't you differentiate the function?

The OP did that in post #1, quoted below.

f'(x)=x(1+sinx)

 

[sergey_le]

The OP did that in post #1, quoted below.

f'(x)=x(1+sinx)

I'm sorry I don't understand what you want to say

 

[Mark44]

I'm sorry I don't understand what you want to say

@PeroK asked you if you could differentiate the function. He must not have noticed that you showed this in post #1.

 

[PeroK]

Also the derivative f'(x)≥0
I don't understand what you want me to do?
How do I show that f'(x)>0 And no ≥

When is $f'(x) = 0$?

What happens when $f'(x) = 0$?

Analyse the derivative!

 

[PeroK]

The OP did that in post #1, quoted below.

f'(x)=x(1+sinx)

Yes, I did miss that. Nevertheless I did say in post #12 that we should start at the beginning!

 

[Mark44]

When is $f'(x) = 0$?
What happens when $f'(x) = 0$?

Yes, and @sergey_le, what are the values of f(x) at the points where f'(x) = 0?

 

[sergey_le]

Yes, and @sergey_le, what are the values of f(x) at the points where f'(x) = 0?

nothing special.
Please direct me

 

[PeroK]

nothing special.
Please direct me

This is painful. Things to try:

1) Sketch a graph of the function.

2) Analyse the derivative.

3) Analyse the points where the derivative is zero: local min, max, saddle point?

4) Look at the second derivative $f''(x)$.

5) Use the fact that $f$ never decreases. If it did $f'(x)$ would be less than zero at some points.

6) What happens for small $x$?

7) Have you noticed the function is symmetric about the y-axis?

You have to be prepared to do some of the thinking here.
7)

 

[Mark44]

nothing special.
Please direct me

I already did.

what are the values of f(x) at the points where f'(x) = 0?

You found f'(x) in post #1 -- f'(x) = x(1 + sin(x)).

Are you saying you don't know how to solve the equation x(1 + sin(x)) = 0?

 

[sergey_le]

I already did.

You found f'(x) in post #1 -- f'(x) = x(1 + sin(x)).

Are you saying you don't know how to solve the equation x(1 + sin(x)) = 0?

You mean I have to show that there is no f' axis with x axis?

 

[Mark44]

You mean I have to show that there is no f' axis with x axis?

I have no idea what that means.
I'm asking whether you can solve the equation x(1 + sin(x)) = 0. This is something you should have done in an algebra/trig class.

 

[sergey_le]

I have no idea what that means.
I'm asking whether you can solve the equation x(1 + sin(x)) = 0. This is something you should have done in an algebra/trig class.

Yes I know how to solve it
x(1 + sin(x)) = 0 when
x=0 or x=3/2+πk

 

[Mark44]

Yes I know how to solve it
x(1 + sin(x)) = 0 when
x=0 or x=3/2+πk

x = 0, yes, but x=3/2+πk is not a solution.

 

[sergey_le]

why x=3/2+πk is not a solution?

 

[PeroK]

why x=3/2+πk is not a solution?

I guess you meant

I guess you meant $x = \frac{3\pi}{2} + 2\pi k$?

So, @sergey_le, now you know that f'(x) = 0 for x = 0 or for $x = \frac{3\pi}{2} + 2\pi k$.
What are the values of f(x) at these x values?

You also know that f'(x) > 0 for $x \ne 0$ and for $x \ne \frac{3\pi}{2} + 2\pi k$.

Remember that you're trying to show that f(x) > 0 for x > 0 as part of the problem.

 

[sergey_le]

I guess you meant $x = \frac{3\pi}{2} + 2\pi k$?

Yes I'm sorry.
I don't know why I make such mistakes

 

[sergey_le]

I guess you meant

So, @sergey_le, now you know that f'(x) = 0 for x = 0 or for $x = \frac{3\pi}{2} + 2\pi k$.
What are the values of f(x) at these x values?

You also know that f'(x) > 0 for $x \ne 0$ and for $x \ne \frac{3\pi}{2} + 2\pi k$.

Remember that you're trying to show that f(x) > 0 for x > 0 as part of the problem.

Thanks so much I finally realized

 

[PeroK]

I thought it might be worth adding an outline solution to this.

We need to show that $f(x) > 0 \$ when $x \ne 0$, where $f(x) = \frac 1 2 x^2 - x\cos x + \sin x$.

Note that $f$ is symmetric about the y-axis ($f(x) = f(-x)$), so it is enough to show that $f(x) > 0$ for $x > 0$. Note also that $f(0) = 0$.

$f'(x) = x + x\sin x = x(1 + \sin x)$

Therefore $f'(x) \ge 0$ for $x > 0$, hence (by the mean value theorem) $f(x) \ge 0$ for $x > 0$.

Note that $f'(x) = 0$ when $x = \frac 3 2 \pi + 2n\pi$. In particular, $f'(x) > 0$ for $x \in (0, \pi]$. Therefore, by the mean value theorem, $f(x) > 0$ for $x \in (0, \pi]$. And, of course, $f(\pi) > 0$.

Finally, for $x > \pi$ we have $f'(x) \ge 0$, hence $f(x) \ge f(\pi) > 0$; and the result follows.