Using the Intermediate Value Theorem on two functions?

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SUMMARY

The discussion focuses on applying the Intermediate Value Theorem (IVT) to two continuous functions, f and g, defined on the interval [a, b]. Given that f(a) < g(a) and g(b) < f(b), the conclusion is that there exists at least one number c in (a, b) such that f(c) = g(c). The proof involves defining a new function h(x) = f(x) - g(x), demonstrating that h(a) < 0 and h(b) > 0, which guarantees the existence of c where h(c) = 0.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem (IVT)
  • Knowledge of continuous functions
  • Ability to manipulate algebraic expressions involving functions
  • Familiarity with function behavior on closed intervals
NEXT STEPS
  • Study the proof of the Intermediate Value Theorem in detail
  • Explore examples of continuous functions and their applications of IVT
  • Learn about the implications of continuity in calculus
  • Investigate the relationship between function intersections and graphical analysis
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Students studying calculus, particularly those learning about the Intermediate Value Theorem and its applications, as well as educators looking for examples to illustrate continuity and function behavior.

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Homework Statement



Given that f and g are continuous on [a, b], that f(a) < g(a), and g(b) < f(b), show that there exists at least one number c in (a, b) such that f(c) = g(c). HINT: Consider f(x) - g(x).

Homework Equations



If f is continuous on [a, b] and K is a number between f(a) and f(b), then there is at lease one number c between a and b for which f(c) = K.

The Attempt at a Solution



I know the proof for the Intermediate Value Theorem (IVT) is as follows:

Suppose F(a) < K < f(b)
g(x) = f(x) - K is continuous on [a, b], since
g(a) = f(a) - K < 0 and g(b) = f(b) - K > 0
We know there is a number c between a and b for which g(c) = 0
Then, f(c) = K

From the hint it seems to say to make a function p(x) = f(x) - g(x) like how g(x) = f(x) - K in the proof. But I don't know where to go from there... it gets confusing with so many functions!
 
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Put h(x)=f(x)-g(x). Then h(a)<0 and h(b)>0. Thus 0\in [h(a),h(b)], which implies that h(c)=0 for some c.
 

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