Prove that this function is nonnegative

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In summary, the conversation discusses how to prove that the function f(x)=##x^2##/2 - xcosx+sinx is greater than 0 for all values of x except 0. The participants suggest using the Mean Value Theorem and a proof by contradiction to show that f(x) is always positive.
  • #1
sergey_le
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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
 
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  • #2
sergey_le said:
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?
 
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  • #3
Mark44 said:
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?
 
  • #4
sergey_le said:
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##.
 
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  • #5
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.
 
  • #6
PeroK said:
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.
 
  • #7
FactChecker said:
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
 
  • #8
sergey_le said:
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##.
 
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  • #9
sergey_le said:
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?
 
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  • #10
PeroK said:
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
 
  • #11
FactChecker said:
Are you allowed to use the Mean Value Theorem?
yes
 
  • #12
sergey_le said:
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?
 
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  • #13
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.
 
  • #14
FactChecker said:
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##.
 
  • #15
FactChecker said:
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
 
  • #16
PeroK said:
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 .
 
  • #17
sergey_le said:
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.
 
  • #18
PeroK said:
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 ≥
 
  • #19
PeroK said:
First, you have to differentiate f(x)f(x)f(x). Can you do that?
PeroK said:
Why don't you differentiate the function?
The OP did that in post #1, quoted below.

f'(x)=x(1+sinx)
 
  • #20
Mark44 said:
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
 
  • #21
sergey_le said:
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.
 
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  • #22
sergey_le said:
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!
 
  • #23
Mark44 said:
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!
 
  • #24
PeroK said:
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?
 
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  • #25
Mark44 said:
Yes, and @sergey_le, what are the values of f(x) at the points where f'(x) = 0?
nothing special.
Please direct me
 
  • #26
sergey_le said:
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)
 
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  • #27
sergey_le said:
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?
 
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  • #28
Mark44 said:
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?
 
  • #29
sergey_le said:
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.
 
  • #30
Mark44 said:
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
 
  • #31
sergey_le said:
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.
 
  • #32
why x=3/2+πk is not a solution?
 
  • #33
sergey_le said:
why x=3/2+πk is not a solution?

I guess you meant
PeroK said:
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.
 
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  • #34
PeroK said:
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
 
  • #35
PeroK said:
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
 
<h2>1. How do you prove that a function is nonnegative?</h2><p>To prove that a function is nonnegative, you need to show that the output of the function is always equal to or greater than zero for all possible inputs. This can be done by using mathematical techniques such as induction, contradiction, or direct proof.</p><h2>2. Can a function be both nonnegative and negative?</h2><p>No, a function cannot be both nonnegative and negative. A function is considered nonnegative if its output is always equal to or greater than zero. If the output is ever less than zero, the function is considered negative.</p><h2>3. What are some common examples of nonnegative functions?</h2><p>Some common examples of nonnegative functions include polynomials, exponential functions, and trigonometric functions. These functions always have a positive or zero output for all possible inputs.</p><h2>4. How does proving a function is nonnegative relate to real-world applications?</h2><p>Proving that a function is nonnegative is important in various real-world applications, such as in economics, physics, and engineering. In economics, nonnegative functions are used to model revenue, profit, and demand. In physics, nonnegative functions are used to represent energy, distance, and time. In engineering, nonnegative functions are used to model signal strength, power, and voltage.</p><h2>5. Can a function be nonnegative on one interval and negative on another?</h2><p>Yes, a function can be nonnegative on one interval and negative on another. This means that the function's output is equal to or greater than zero for some values of the input, but less than zero for other values of the input. In this case, the function is considered nonnegative only on the interval where its output is always equal to or greater than zero.</p>

1. How do you prove that a function is nonnegative?

To prove that a function is nonnegative, you need to show that the output of the function is always equal to or greater than zero for all possible inputs. This can be done by using mathematical techniques such as induction, contradiction, or direct proof.

2. Can a function be both nonnegative and negative?

No, a function cannot be both nonnegative and negative. A function is considered nonnegative if its output is always equal to or greater than zero. If the output is ever less than zero, the function is considered negative.

3. What are some common examples of nonnegative functions?

Some common examples of nonnegative functions include polynomials, exponential functions, and trigonometric functions. These functions always have a positive or zero output for all possible inputs.

4. How does proving a function is nonnegative relate to real-world applications?

Proving that a function is nonnegative is important in various real-world applications, such as in economics, physics, and engineering. In economics, nonnegative functions are used to model revenue, profit, and demand. In physics, nonnegative functions are used to represent energy, distance, and time. In engineering, nonnegative functions are used to model signal strength, power, and voltage.

5. Can a function be nonnegative on one interval and negative on another?

Yes, a function can be nonnegative on one interval and negative on another. This means that the function's output is equal to or greater than zero for some values of the input, but less than zero for other values of the input. In this case, the function is considered nonnegative only on the interval where its output is always equal to or greater than zero.

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