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I am self-studying Calculus and tried to solve the following question:

Suppose that the function f is continuous and increasing in the closed interval [a, b]. Then

(i) f has an inverse f

(ii) f

(iii) f

The book does the proofs for (i) and (ii). It then proofs (iii) partially by showing that f is continuous in the open interval (a,b).

I must then show that f

Definition of unilateral limit to the right:

[tex]\lim_{x\to a^+} f(x) = L[/tex]

if and only if, for every small ε > 0, there is a δ > 0 such that

|f(x) - L| < ε whenever 0 < x - a < δ

I will try to do here the proof that f

Applying the definition of unilateral limit to the right, I must then show that, for every small ε > 0, there is a δ > 0 such that

|f

I begin by defining ε such that a + ε belongs to [f(a), f(b)].

Since a < a + ε, then, since f is increasing:

[tex]f(a) < f(a+\epsilon)[/tex]

Now, since the choice of δ depends upon the choice of ε, I choose δ such that:

[tex]\delta \leq f(a+\epsilon) - f(a)[/tex]

[tex]f(a)+\delta \leq f(a+\epsilon)[/tex]

Now, from the definition of the limit above, I know that:

[tex]0 < y - f(a) < \delta[/tex]

Summing f(a) to every term:

[tex]f(a) < y < f(a) + \delta[/tex]

So, combining this with the δ chosen above:

[tex]f(a) < y < f(a) + \delta \leq f(a+\epsilon)[/tex]

Since f and f

[tex]f^{-1}(f(a)) < f^{-1}(y) < f^{-1}(f(a+\epsilon))[/tex]

[tex]a < f^{-1}(y) < a+\epsilon[/tex]

Subtracting a from every term:

[itex]0 < f^{-1}(y) - a < \epsilon[/itex] whenever [itex]0 < y - f(a) < \delta[/itex]

which appears to be the desired result, except for the lack of absolute value. I imagine the absolute value didn't appear here because [itex]f^{-1}[/itex] is an increasing function.

Thank you in advance.

## Homework Statement

Suppose that the function f is continuous and increasing in the closed interval [a, b]. Then

(i) f has an inverse f

^{-1}, which is defined in [f(a), f(b)];(ii) f

^{-1}is increasing in [f(a), f(b)];(iii) f

^{-1}is continuous in [f(a), f(b)].The book does the proofs for (i) and (ii). It then proofs (iii) partially by showing that f is continuous in the open interval (a,b).

I must then show that f

^{-1}is continuous in f(a) to the right and continuous in f(b) to the left.## Homework Equations

Definition of unilateral limit to the right:

[tex]\lim_{x\to a^+} f(x) = L[/tex]

if and only if, for every small ε > 0, there is a δ > 0 such that

|f(x) - L| < ε whenever 0 < x - a < δ

## The Attempt at a Solution

I will try to do here the proof that f

^{-1}is continuous in f(a) to the right.Applying the definition of unilateral limit to the right, I must then show that, for every small ε > 0, there is a δ > 0 such that

|f

^{-1}(y) - a| < ε whenever 0 < y - f(a) < δI begin by defining ε such that a + ε belongs to [f(a), f(b)].

Since a < a + ε, then, since f is increasing:

[tex]f(a) < f(a+\epsilon)[/tex]

Now, since the choice of δ depends upon the choice of ε, I choose δ such that:

[tex]\delta \leq f(a+\epsilon) - f(a)[/tex]

[tex]f(a)+\delta \leq f(a+\epsilon)[/tex]

Now, from the definition of the limit above, I know that:

[tex]0 < y - f(a) < \delta[/tex]

Summing f(a) to every term:

[tex]f(a) < y < f(a) + \delta[/tex]

So, combining this with the δ chosen above:

[tex]f(a) < y < f(a) + \delta \leq f(a+\epsilon)[/tex]

Since f and f

^{-1}are both increasing:[tex]f^{-1}(f(a)) < f^{-1}(y) < f^{-1}(f(a+\epsilon))[/tex]

[tex]a < f^{-1}(y) < a+\epsilon[/tex]

Subtracting a from every term:

[itex]0 < f^{-1}(y) - a < \epsilon[/itex] whenever [itex]0 < y - f(a) < \delta[/itex]

which appears to be the desired result, except for the lack of absolute value. I imagine the absolute value didn't appear here because [itex]f^{-1}[/itex] is an increasing function.

Thank you in advance.

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