Prove that the function in defined

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SUMMARY

The function f(x) is defined as f(x) = 1/((ln(x+1))^2 + 1) for x > -1 and f(x) = 0 for x = -1. The integral F(x) = ∫(from 0 to x^2 + 2x) f(t) dt is defined for all real numbers R and possesses a derivative due to the continuity of f(x) for x ≥ -1. Additionally, the function g(x) is defined as g(x) = f(x) for x > -1 and g(x) = -1 for x = -1, leading to G(x) = ∫(from 0 to x^2 + 2x) g(t) dt, which is also defined in R and has a derivative. The change in g(x) at x = -1 does not affect the integral due to the measure zero property of the set {-1}.

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Homework Statement



f(x) is defined as f(x) = 1/((ln(x+1))^2 + 1) for all x>-1 and f(x)=0 for x=-1.

1)Prove that the function [tex]F(x) = \int^{x^2 + 2x}_{0} f(t)dt[/tex]
is defined and has a derivative in R.
2)g(x) is defined as g(x)=f(x) for x>-1 and g(x)=-1 for x=-1.
Also, [tex]G(x) = \int^{x^2 + 2x}_{0} g(t)dt[/tex]
Is G(x) defined in R? Does it have a derivative?

Homework Equations





The Attempt at a Solution



1) By taking the limit of f(x) at x=0 we see that f is continues for all x>=1 and since
x^2 + 2x >= -1 for all x in R F(x) is defined and is has a derivative from the chain rule.

2)Since f(x)=g(x) for all x=/=-1 F(x)=G(x) and so the answer to both questions is yes.

Are those right? I think that the answer to (2) is wrong but why?
Thanks.
 
Last edited:
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Seems fine to me. For 2), why would you think changing the value of a function at a single point could change the integral? The set {-1} has measure 0.
 
Thanks a lot.
 

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