Finding a function, given derivative

In summary: I'm confused, I don't see how you can find r(x) when you are given s(x).The problem I'm working on is to find s(x).How do I find r(x)?I'm confused, I don't see how you can find r(x) when you are given s(x).The problem I'm working on is to find s(x).How do I find r(x)?What you have to do is find a function r(x) that satisfies the following properties:1) r(0) = 02) r(x) > s(x) for all x (except possibly at x = 0)3) r'(x) > s'(x) for all
  • #1
Maybe_Memorie
353
0

Homework Statement



Prove that there is a function s, defined on all of R, such that

s(0) = 0 and s′(x) = (1 + x^4)^(−1/2)

Show that s is bounded.

Homework Equations



Integration by parts

The Attempt at a Solution



Right, I tried using integration by parts. That didn't get me anywhere.

A hint to get me started would be very helpful. Thanks :)
 
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  • #2
Maybe_Memorie said:

Homework Statement



Prove that there is a function s, defined on all of R, such that

s(0) = 0 and s′(x) = (1 + x^4)^(−1/2)

Show that s is bounded.

Homework Equations



Integration by parts

The Attempt at a Solution



Right, I tried using integration by parts. That didn't get me anywhere.

A hint to get me started would be very helpful. Thanks :)

I think that you are misreading this problem. It doesn't ask you to find s(x) - it asks you to show that s(x) is bounded.
 
  • #3
@Mark : What other way can he use to prove that 's' exists besides integration? It is given that after proving that s(x) exists, to show that it is bounded.
The integral isn't an elementary one though...can't be done using Riemann integrals.
 
  • #4
This problem looks to me like it might be an application of the Fundamental Theorem of Calculus.
[tex]\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}[/tex]

From this it's easy to see that s(0) = 0 and s'(x) = 1/sqrt(1 + x^4).

I haven't worked it through, but this is what I'd start with.
 
  • #5
Find a function f(x) such that f(x) > s'(x) for all real x and such that: [tex]\int_0^\infty f(x)\,dx[/tex] converges. Then s(x) is bounded above.

This works because s'(x) > 0 for all real x.

Show that s(x) is an odd function to get a lower bound.

Hint to help find such a function, f :

[tex]\frac{1}{x^2}>\frac{1}{\sqrt{1+x^4}}\quad\text{and}\quad\int_{a}^\infty\frac{1}{x^2}\,dx=\frac{1}{a}\ \ \text{ for }\ \ a>0\,.[/tex]

A piecewise function will do the trick.
 
  • #6
Mark44 said:
This problem looks to me like it might be an application of the Fundamental Theorem of Calculus.
[tex]\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}[/tex]

From this it's easy to see that s(0) = 0 and s'(x) = 1/sqrt(1 + x^4).

I haven't worked it through, but this is what I'd start with.


I'm kind of lost as to what to do next
 
  • #7
Define: r'(x)=min(1, 1/x2), and r(0)=0.

Show that r'(x) ≥ s'(x) for all real x, in fact, r'(x) > s'(x) except for x = 0 at which they're equal.

r(x) > s(x) for x > 0.

Use integration to find r(x).

Both r'(x) & s'(x) are even functions which take on only positive values.

Both r(x) & s(x) are odd functions and monotonic increasing. (Why?)

Toss in a few other details, & you're done.
 
  • #8
I found s(x) and showed that it was unique.

I'm not sure how to show that it is bounded though
 
  • #9
Maybe_Memorie said:
I found s(x) and showed that it was unique.

I'm not sure how to show that it is bounded though
What did you get for s(x)?

s'(x) > 0, so that says that s(x) is monotonic increasing.

If you can determine the limits, limx→±∞ s(x), that should give you the bounds.

If you use the function r'(x) from my previous post, that will also give you bounds, just not the least upper bound or greatest lower bound.
 
  • #10
well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0
 
  • #11
I noticed the question never actually said "Find s(x)". All I had to do was show it exists, is defined on all of R, is unique, and is bounded.

[tex]\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}[/tex]

s(0) clearly equals 0 and s(x) is clearly defined on all of R.

Suppose f(x) also has the desired properties.
Then f'(x) = s'(x)
So f(x) = s(x) + C , C is a constant.

s(0) = 0 = f(0) therefore C = 0.
Then f(x) = s(x) and the function is unique.


Now how do I show it is bounded?
 
  • #12
judowrestler1 said:
well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0
?
If F is defined by
[tex]F(x)= \int_0^x f(t)dt[/tex]
for any integrable function f, then F(0)= 0.
 
  • #13
You know the derivative, my first impulse would be to find out the values of the function s'(x) at some various points, say [tex]0,\pm\infty[/tex].
 
  • #14
hunt_mat said:
You know the derivative, my first impulse would be to find out the values of the function s'(x) at some various points, say [tex]0,\pm\infty[/tex].

x = 0, s'(x) = 1
x = plus or minus infinity, s'(x) = 0
 
  • #15
judowrestler1 said:
well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0
Not true.

d[ln((x^2)+sqrt(1+(x^4)))]/dx = 2x/√(1+x4) ≠ s'(x)
 
  • #16
Maybe_Memorie said:
I noticed the question never actually said "Find s(x)". All I had to do was show it exists, is defined on all of R, is unique, and is bounded.

[tex]\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}[/tex]

s(0) clearly equals 0 and s(x) is clearly defined on all of R.

Suppose f(x) also has the desired properties.
Then f'(x) = s'(x)
So f(x) = s(x) + C , C is a constant.

s(0) = 0 = f(0) therefore C = 0.
Then f(x) = s(x) and the function is unique.Now how do I show it is bounded?
Use the comparison test with s'(x) and the r'(x) I suggested.

Another way to describe r'(x) is:

[tex]r'(x)=\left\{\begin{array}{cc}1,&\mbox{ if }
-1\leq x\leq 1\\ \displaystyle \frac{1}{x^2}\ \ , & \mbox{ otherwise } \end{array}\right.[/tex]

Compare the graphs of r'(x) and s'(x) . This may help you see how finding r(x) can help show that s(x) is bounded.
 
  • #17
Here is a link to a plot by WolframAlpha showing the two functions: r'(x) and s'(x), where r'(x) is defined as:

[tex]r'(x)=\left\{\begin{array}{cc}1,&\mbox{ if }
-1\leq x\leq 1\\ \\ \displaystyle \frac{1}{x^2}\ \ , & \mbox{ otherwise } \end{array}\right.[/tex]

The function, r(x), where r(0) = 0, is easily found by integration:

[tex]r(x) = \int_0^x r'(t)\,dt[/tex]

[tex]r(x)=\left\{\begin{array}{cc}x,&\mbox{ if }
-1\leq x\leq 1\\ \\ \displaystyle 2-\frac{1}{x}\ \ , & \mbox{ if }
x>1 \\ \\ \displaystyle -2-\frac{1}{x}\ \ , & \mbox{ if }
x<-1 \end{array}\right.[/tex]

.
 
Last edited:

1. What is the general process for finding a function given its derivative?

The general process for finding a function given its derivative is to integrate the derivative with respect to the independent variable. This will result in the original function, up to a constant of integration. In other words, if the derivative is f'(x), the original function is f(x) = ∫ f'(x) dx + C, where C is the constant of integration.

2. Can a function have more than one derivative?

Yes, a function can have an infinite number of derivatives. Each successive derivative represents the slope of the tangent line at a given point on the graph of the original function. Therefore, the first derivative represents the rate of change of the original function, the second derivative represents the rate of change of the first derivative, and so on.

3. What is the relationship between a function and its derivative?

The derivative of a function represents the instantaneous rate of change of that function at any given point. In other words, it shows how much the function is changing at a specific point. The derivative is also equal to the slope of the tangent line at that point on the graph of the function.

4. Can a function's derivative ever be undefined?

Yes, a function's derivative can be undefined at certain points on its graph. This occurs when the function is not continuous or differentiable at that point. For example, a function with a sharp corner or a vertical tangent line will have an undefined derivative at that point.

5. How can I use the derivative to find the maximum or minimum value of a function?

The maximum or minimum value of a function can be found by setting its derivative equal to 0 and solving for the independent variable. This will give the x-coordinate of the point where the function has a maximum or minimum value. To determine whether it is a maximum or minimum, you can use the second derivative test, which involves evaluating the second derivative at that point. If the second derivative is positive, the point is a minimum, and if it is negative, the point is a maximum.

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