Help with multivariate Calculus

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SUMMARY

The discussion focuses on proving the equation max{a,b} = 1/2*(a+b+|a-b|) and demonstrating that if f, g: R → R are continuous functions, then h(x) = max{f(x), g(x)} is also continuous. Additionally, it addresses the behavior of the function g(x) = f(x + 1 - f(x)) as x approaches infinity, given that f: (0, ∞) → R is differentiable and f(x) approaches 0. The solution approach involves analyzing two cases based on the relationship between a and b.

PREREQUISITES
  • Understanding of basic calculus concepts, including limits and continuity.
  • Familiarity with differentiable functions and their properties.
  • Knowledge of absolute value functions and their implications in mathematical proofs.
  • Experience with piecewise functions and their continuity.
NEXT STEPS
  • Study the properties of continuous functions and their implications in calculus.
  • Learn about piecewise-defined functions and how to analyze their continuity.
  • Explore the concept of limits and how they apply to differentiable functions.
  • Practice solving problems involving max functions and absolute values in calculus.
USEFUL FOR

Students preparing for multivariate calculus, educators teaching calculus concepts, and anyone interested in understanding the continuity of functions and their properties.

am100181
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Homework Statement



Show that:

max{a,b}=1/2*(a+b+|a-b|)

Hence or otherwise, show that if f,g:R ----> R are continuous, then the function h:R ----> R defined by h(x) = max{f(x),g(x)}
is also continuos.

Suppose the f:(0,inf) ---> R is differentiable and f(x) ---> 0 as x---> inf. Set g(x) = f(x+1-f(x). Prove that g(x) ---->0 and x----> inf.


The Attempt at a Solution



How do I start attempting such questions? I am thinking of taking multivariate next sem and am wanting to prepare myself for such questions...

Thank you
 
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Welcome to PF!

am100181 said:
Show that:

max{a,b}=1/2*(a+b+|a-b|)

How do I start attempting such questions?

Hi am100181! Welcome to PF! :wink:

Just treat the two cases a≥ b and a< b separately. :smile:
 

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