Antiderivative of 4-3(1+x^2)^-1 | Graph Comparison

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Homework Help Overview

The problem involves finding the antiderivative of the function f(x) = 4 - 3(1 + x^2)^-1, with the condition that F(1) = 0. Participants are tasked with checking their work by comparing the graphs of f and F.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of the antiderivative F(x) and express uncertainty about the correctness of their expressions. There is an attempt to derive the constant C by substituting the condition F(1) = 0 into their expressions.

Discussion Status

Some participants are seeking confirmation on their derived expressions for F(x) and are exploring the implications of the condition F(1) = 0. There is a mix of attempts to clarify the differentiation of their expressions and to solve for the constant C.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring the correctness of their antiderivative and the associated conditions.

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Homework Statement


Find the antiderivative F of f that satisifies the given condition. Check by comparing the graphs of f and F.

f(x)=4-3(1+x^2)^-1
F(1)=0


Homework Equations





The Attempt at a Solution


So far what I attempted to do was this:

F(x)=4x -3tan^-1(x) + C
but was unsure from here on.
 
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Just plug in the given value for F(1) = 0 and solve for C.
 
Ok so my F(x) is correct?

That was probably the part I needed the most assurance with.
 
master1425 said:

Homework Statement


Find the antiderivative F of f that satisifies the given condition. Check by comparing the graphs of f and F.

f(x)=4-3(1+x^2)^-1
F(1)=0


Homework Equations





The Attempt at a Solution


So far what I attempted to do was this:

F(x)=4x -3tan^-1(x) + C
but was unsure from here on.
(4x)'= 4 and
(-3 arctan(x))'= -3/(x2+1)
so what is F' ?

F(1)= 4(1)- 3arctan(1)+ C= 0. C= ?
 

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