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

In summary, the problem is asking to show that if a function g(x) is continuous and differentiable on the interval [a,b] and g'(x) is never equal to 0 on that interval, then g(x) is injective on that interval. This means that for any two different values of x, the corresponding values of g(x) will also be different. It can be proven using the definition of injective and Rolle's theorem, which states that if a function is continuous and differentiable on an interval and has the same values at the endpoints, then there exists a point within that interval where the derivative is equal to 0.
  • #1
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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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  • #2
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
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a mathematical theorem that states that for a given function on a closed interval, there exists a point within that interval where the slope of the tangent line is equal to the average rate of change of the function.

2. What is the significance of proving injectivity on a closed interval?

Proving injectivity on a closed interval using the Mean Value Theorem is important because it allows us to show that a function is one-to-one, or that each input has a unique output. This can help us analyze the behavior of a function and make predictions about its outputs.

3. How is the Mean Value Theorem used to prove injectivity?

The Mean Value Theorem is used to prove injectivity by showing that if a function is not one-to-one, then there must be two distinct points in the interval with the same slope of the tangent line. This contradicts the Mean Value Theorem, which states that there can only be one point with that property.

4. What are the prerequisites for proving injectivity using the Mean Value Theorem?

In order to prove injectivity using the Mean Value Theorem, one must have a good understanding of the concept of derivatives, as well as the definition of a one-to-one function. It is also important to have a thorough understanding of the Mean Value Theorem itself and its various forms.

5. Can the Mean Value Theorem be used to prove injectivity for all functions on a closed interval?

No, the Mean Value Theorem can only be used to prove injectivity for continuous functions on a closed interval. If a function is not continuous, then the Mean Value Theorem cannot be applied. Additionally, there may be other ways to prove injectivity for certain functions that do not involve the Mean Value Theorem.

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