Prove: Increasing Function f(x) Has a Positive Value at a

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SUMMARY

The discussion centers on proving that an increasing and concave up function f(x) must have a point "a" where f(a) > 0. The proof begins by selecting an arbitrary point x0 in the function's domain. If f(x0) is not positive, the increasing nature of f(x) implies that there exists a point x > x0 where f(x) must exceed zero, thus establishing the existence of such an "a". This conclusion leverages the definitions of increasing functions and concavity in relation to derivatives.

PREREQUISITES
  • Understanding of increasing functions and their properties
  • Knowledge of concave up functions and their implications
  • Familiarity with derivatives and their role in function behavior
  • Basic proof techniques in calculus
NEXT STEPS
  • Study the definitions and properties of increasing functions in calculus
  • Learn about concavity and its implications on function behavior
  • Explore proof techniques involving the Intermediate Value Theorem
  • Investigate the relationship between derivatives and function growth
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Students studying calculus, mathematicians interested in function analysis, and educators teaching concepts of increasing and concave functions.

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Homework Statement



Suppose that a function f(x) is increasing and concave up. Show that there is a number "a", such that f(a)>0.

The Attempt at a Solution



How would I go about showing this?
 
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What do "increasing" and "concave up" mean (in terms of derivatives)?

I think the best way to start the proof is: Pick an arbitrary x0 in the domain.
Then either f(x0) > 0 and you are done, or f(x0) <= 0. Now probably your intuition tells you that somewhere to the right of x0, f must become positive, right?
You can formalise that by taking some x > x0, and making estimates for f(x) in terms of f(x0) and positive quantities, until you find an x = a such that f(a) > 0.
 

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