# Prove that this function is nonnegative

## Homework Statement:

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

## Relevant 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,∞)
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

## Answers and Replies

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Mark44
Mentor
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.
sergey_le said:
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?

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
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Homework Helper
Gold Member
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
Science Advisor
Gold Member
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.

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.

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
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Homework Helper
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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 $z > 0$.

Last edited:
FactChecker
Science Advisor
Gold Member
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?

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

Are you allowed to use the Mean Value Theorem?
yes

PeroK
Science Advisor
Homework Helper
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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
Science Advisor
Gold Member
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
Science Advisor
Homework Helper
Gold Member
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 $f(x) > 0$.

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

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
Science Advisor
Homework Helper
Gold Member
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.

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
Mentor
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)

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
Mentor
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
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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
Science Advisor
Homework Helper
Gold Member
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
Mentor
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?

Yes, and @sergey_le, what are the values of f(x) at the points where f'(x) = 0?
nothing special.
Please direct me