Proving the Derivative Function for a Differentiable Function on an Interval

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SUMMARY

The discussion centers on proving the derivative function for a differentiable function defined on the interval [0, b]. The average value of the function f(x) is given by (f(0) + f(b))/2 for b > 0. The integral of f(x) from 0 to 3 is calculated to be 6. The proof of the derivative f'(x) is established as (f(x) - 5)/x for all x < 0, utilizing the average value formula and differentiation techniques.

PREREQUISITES
  • Differentiation techniques
  • Understanding of average value of a function
  • Basic integral calculus
  • Knowledge of the Mean Value Theorem
NEXT STEPS
  • Study the Mean Value Theorem in calculus
  • Learn about differentiation of implicit functions
  • Explore the properties of definite integrals
  • Review the concept of average value of a function over an interval
USEFUL FOR

Students studying calculus, particularly those focusing on differentiation and integration concepts, as well as educators looking to enhance their understanding of derivative proofs.

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Homework Statement


Let f be a differentiable function defined for all x>=0 such that f(0)=5 and f(3)=-1. Suppose that for any number b>0, the average value of f(x) on the interval 0<=x<=b is (f(0)+f(b))/x

a. Find the integral of f(x) from 0 to 3.
b. Prove that f'(x)=(f(x)-5)/x for all x<0.
c. Using part b), find f(x)

Homework Equations


(b-a)(f((ave)x))= the integral of f(x) from a to b

The Attempt at a Solution


Part a is easy, I got 6 as my answer. I'm completely at a loss on how to do part b/c. If anyone would at least point me in the right direction, I would greatly appreciate it.
 
Last edited:
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Suppose that for any number b>0, the average value of f(x) on the interval 0<=x<=b is (f(0)+f(b))/x
Either you copied out the question wrong or the question asked doesn't make sense.
 
Argh, yes I did copy it wrong. It should be (f(0)+f(b))/2.
 
Use your "relevant equations" together with the given formula for the average of f over [0,b] to get an equation. Differentiate both sides of the equation.
 

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