The discussion centers on the geometric derivation of the derivative of the integral function f(x) = ∫(^x, _0) (dt)/(1+t^2), which is claimed to equal arctan(x). However, participants clarify that the derivative is actually f'(x) = 1/(1+x^2), not arctan(x). They suggest using the difference quotient limit definition of the derivative and the Mean Value Theorem for integrals to establish the relationship geometrically. The squeeze theorem is also recommended as a method to bound the integral and find the limit.
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Laurentiusson said:It looks like my A is 1 and my B is 0 since the graph has a max at 1 and approaches 0 both to left and right.
Mark44 said:No. A and B aren't the max and min values of 1/(1 + t2). They are the max and min values of the integral.
I agree. We can't go further than the hints we have given without working the problem for you, which as LCKurtz says, is against the policy in this forum.LCKurtz said:After participating in and watching this thread, I don't think we can get you through this problem without working it for you, which is against forum policy. My suggestion is that it is time for you to have a face to face meeting with your teacher.
If you have a specific question about something we have written, ask it, but saying you are more and more confused doesn't give anything to go on.Laurentiusson said:I'm just more and more confused for every sentence you write. What am I actually suppoadd to do?
On the interval [x, x + h], what is the maximum value of 1/(1 + t2)? What is the minimum value of 1/(1 + t2)?Laurentiusson said:I don't have a teacher to ask which is the reason to why I'm here.
I wonder how I can see the max and mins for 1/(1+t^2). I thought I just should look for extreme points when I plot it but I guess I'm wrong.