Prove a composite function is increasing

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SUMMARY

The discussion centers on proving that the function g(x) = f(x)/x is increasing for all x > 0, given that f is continuous for x ≥ 0 and differentiable for x > 0, with f(0) = 0 and f' being increasing. The user successfully applies the Quotient Rule to derive g' = [xf'(x) - f(x)]/x² and establishes that g is increasing if xf'(x) - f(x) ≥ 0. The conclusion hinges on the Mean Value Theorem, which confirms the existence of a point c in (0, x) such that f(c) = (f(x) - f(0))/(x - 0), aiding in the proof.

PREREQUISITES
  • Understanding of calculus concepts such as derivatives and the Quotient Rule.
  • Familiarity with the Mean Value Theorem and its implications.
  • Knowledge of continuous and differentiable functions.
  • Basic limit evaluation techniques in calculus.
NEXT STEPS
  • Study the application of the Quotient Rule in calculus.
  • Learn about the Mean Value Theorem and its proofs.
  • Explore the properties of increasing functions and their derivatives.
  • Review limit evaluation techniques, particularly in the context of derivatives.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and function behavior, as well as educators looking for examples of applying the Mean Value Theorem in proofs.

Salzburg
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Homework Statement


Hi,
I have trouble proving this claim and would really appreciate your help =). Thank you in advance!
So here's the question: Suppose that f is a continuous function for all x>= 0 and differentiable for all x> 0. Also, f(0) = 0 and f' (1st derivative of f) is increasing on its domain Define
g(x) = f(x)/ x for x >0.
Show that g is increasing for all x >0.

The Attempt at a Solution


I used the Quotion Rule to write g' = [xf'(x) - f(x)]/ x^2.
For g to be increasing, g' must be >= 0 --> the question becomes: proving xf'(x) - f(x) >= 0.
Since f' is increasing -> f'(x) > = f(0) with x >0
--> It is sufficient to have xf'(0) - f(x) >0
I wrote f'(0): by definition of derivative:
f'(0) = lim [f(x) - f(0)]/ x-0 for x --> 0
= lim f(x)/x
Here I got stuck with the limit; it does not allow me to cross-multiple the terms.
 
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so rearranging it remains to show that
[tex]f'(x) > \frac{f(x)}{x}[/tex]

continuous & differentiable work with the mean value theorem, so there exists 0<c<x such that
[tex]f(c) = \frac{f(x) -f(0)}{x-0}[/tex]
 
Thanks for answering my question =). I couldn't believe that I didn't think about the Mean Value Theorem...
 

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