Finding Concavity of y = Integral from x to 0

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Homework Help Overview

The discussion revolves around determining the intervals of concavity for the function defined by the integral \( y = \int_x^0 \frac{1}{1 + t + t^2} dt \). Participants are exploring the conditions under which the function is concave upward.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the general condition for concavity, specifically referencing the second derivative test. There are attempts to differentiate the integral function and find its first and second derivatives. Questions arise about the correctness of these derivatives and how to handle the resulting inequalities.

Discussion Status

There is an ongoing exploration of the derivatives of the function, with some participants providing guidance on the differentiation process. Multiple interpretations of the differentiation steps are being examined, and there is no explicit consensus on the correctness of the derivatives yet.

Contextual Notes

Participants express uncertainty regarding the complexity of the inequalities involved and the application of the Fundamental Theorem of Calculus in this context. There is a noted lack of textbook references during the discussion.

powp
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Hello
How do I do the following
Find the interval on which the curve
[tex]y = \int_x^0 \frac{1}{1 + t + t^2} dt[/tex]
is concave upward.

any help would be great.

P

Just in case Latex does not show up
x
Large S 1/1(1 + t + t^2) dt
0
 
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When, in general, is some function f concave upwards?
 
I believe a function f should be concave upward when f''(c) > 0. Am I correct. Don't have my textbook on me.
 
powp said:
I believe a function f should be concave upward when f''(c) > 0. Am I correct. Don't have my textbook on me.
Correct.

Now, if the function f (t) is [tex]y = \int_x^0 \frac{1}{1 + t + t^2} dt[/tex], then what's f'(t). (This is very easy).

Finding f''(t) is a little harder. Once found set it up as an inequality and solve.
 
f' is (x)[tex]y = \int_x^0 \frac{-1-2t}{(1 + t + t^2)^2}[/tex]

f'' would then be
[tex]y = \int_x^0 \frac{-1 + 2t + 2t^2}{(1 + t + t^2)^4} dt[/tex]

is this correct??

How do I solve an inequality that is this complex(it is complex to me)?? I am really not sure about this.
 
powp said:
f' is (x)[tex]y = \int_x^0 \frac{-1-2t}{(1 + t + t^2)^2}[/tex]
f'' would then be
[tex]y = \int_x^0 \frac{-1 + 2t + 2t^2}{(1 + t + t^2)^4} dt[/tex]
is this correct??
How do I solve an inequality that is this complex(it is complex to me)?? I am really not sure about this.
The integral is also called the anti-derivative, so the derivative of [tex]y = \int_x^0 \frac{1}{1 + t + t^2} dt[/tex]
is just:
[tex]f'(t)= \frac{1}{1 + t + t^2}[/tex]

So your second derivative is just:
[tex]f''(t) =\frac{-1-2t}{(1 + t + t^2)^2}[/tex]
 
Not quite BobG, think about where the x is and what your actually differentiating.
 
anybody with some guidence?
 
powp said:
anybody with some guidence?
Erm, you've been given loads and loads, what have you done with what you have been given?
 
  • #10
"Fundamental Theorem of Calculus"!


What is the derivative of [tex]\int_a^x f(t)dt[/tex] according to the Fundamenta Theorem of Calculus? (This is what Bobg was doing.)

Knowing that, what is the derivative of [tex]\int_x^a f(t)dt[/tex]? (This is what Bobg should have done!)


What is the derivative of [tex]\int_x^a f(t)dt[/tex]
 

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