Register to reply

Do derivatives belong to unique functions?

by rhuelu
Tags: belong, derivatives, functions, unique
Share this thread:
Sep13-08, 06:26 PM
P: 17
I need to show that f(x)=cexp(-cx) is the only solution to c-[integral from 0 to x](f(x)) = f(x). Is this trivial??? If not how would you suggest I go about showing that functions have unique derivatives (after taking into account constants). Thanks in advance for suggestions!
Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Sep13-08, 07:27 PM
Sci Advisor
HW Helper
P: 2,020
Is that equation really

[tex]f(x) = c - \int_0^x f(t) \, dt?[/tex]

Because f(x)=cexp(-cx) isn't a solution to this for all c.
Sep14-08, 11:53 AM
P: 17
the integral should be multiplied by c as well. This is actually a stats problem...the entire question is to show when the hazard/failure rate is constant -> c=f(x) / (1-F(x)) where f=pdf, F=cdf. Its obvious that f(x)=cexp(-cx) is a solution to this. I just don't know how to show its the only solution.

Sep14-08, 12:12 PM
Sci Advisor
PF Gold
P: 39,565
Do derivatives belong to unique functions?

The fact that "derivatives belong to unique functions" or, more correctly, that if f(x) and g(x) have the same derivative then f(x) and g(x) differ at most by a constant, is given in most calculus books and follows from the mean value theorem:

Lemma: Suppose f(x) is continuous on [a, b], is differentiable on (a, b), and f'(x)= 0 for all x in (a, b). Then f(x)= C (f(x) is a constant) for all x in (a,b).

Let x be any point in (a, b). By the mean value theorem, (f(x)- f(a))/(x- a)= f '(c) for some c between a and x. Since f'= 0 between a and b, f'(0)= 0 from which it follows that f(x)- f(a)= 0 or f(x)= f(a). That is, f(x) is equal to the number f(a) for all x between a and b and so f(x) is a constant there.

Theorem: If f(x) and g(x) are both continuous on [a, b], differentiable on (a, b), and f'(x)= g'(x) for all x in (a, b), then f(x)= g(x)+ C, a constant.

Let H(x)= f(x)- g(x). Then H(x) is continuous on [a,b] and differentiable on (a, b). Further, for all x in (a, b), H'(x)= f'(x)- g'(x)= 0 because f'(x)= g'(x). By the lemma, H(x)= f(x)+ g(x)= C for some number C. Then f(x)= g(x)+ C.
Sep14-08, 12:55 PM
P: 17
thanks...that makes sense. Do you think this logic still holds if we only know that F(x) is right continuous? This is one of the properties of a cdf.

Register to reply

Related Discussions
Derivatives of ln functions Calculus & Beyond Homework 6
Derivatives of vector functions Calculus & Beyond Homework 2
Derivatives of Trigonometric Functions Calculus & Beyond Homework 3
Derivatives of Trigonometric Functions Calculus & Beyond Homework 2
Derivatives of Composite Functions Calculus & Beyond Homework 4