Mean Value Theorem: Prove Injective on [a,b]

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SUMMARY

The discussion centers on proving that a function g: [a,b] → R is injective if its derivative g'(x) is non-zero for all x in the interval (a,b). The key concept is that injectivity, or being one-to-one, means that if g(x) = g(y), then x must equal y. The absence of turning points or zero gradients in the interval indicates that g cannot take the same value at two different points, thus confirming its injective nature. The discussion references Rolle's theorem to clarify the implications of continuity and differentiability on the behavior of g.

PREREQUISITES
  • Understanding of the Mean Value Theorem
  • Knowledge of Rolle's Theorem
  • Familiarity with concepts of injective functions
  • Basic calculus, specifically derivatives and their implications
NEXT STEPS
  • Study the Mean Value Theorem and its applications in proving function properties
  • Explore examples of injective and non-injective functions
  • Learn about the implications of continuity and differentiability in calculus
  • Review the definitions and proofs related to Rolle's Theorem
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Students studying calculus, particularly those focusing on function properties and theorems related to continuity and differentiability. This discussion is beneficial for anyone looking to understand injective functions and their proofs in mathematical analysis.

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Homework Statement


Let a>b be Real numbers and
f, g: [a,b] --> R be continuous and differentiable on (a,b)

Show g is injective on [a,b] if g'(x) != 0 for all x in (a,b)

Homework Equations


Rolle's theorem: Continuity and differentiability (in the conditions above) imply that
f(a) = f(b) and there exists c in (a,b) such that f'(c) = 0

The Attempt at a Solution


Well first I don't know exactly what injective means (what is "distinctness"). What I do understand is Rolle's theorem: that there is a turning point or point of zero gradient between any two points that have the same y-value (if that's right). So in this question there is no turning point or zero gradient in the interval [a,b] - but I don't know what the function is restricted to look like. I'm thinking it could be a horizontal straight line, a parabola, or a squiggly thing that starts and ends between two horizontal points. I'm really quite clueless how to prove something for all situations

If you could just give me a starting point or outline,
Thanks
 
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You may have heard the term 1-1 before.

The definition for injective (1-1 is the same thing) is: if g(x)=g(y) then x=y. For example, the function y=x2 is not injective because (-1)2=12, but the function y=x3 is injective because if a3=b3 this forces a=b

In fact in your post you almost stated a solution to the problem: a re-wording of this definition is that g(x) is not injective if there are two different values of x for which g(x) is the same
 

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