4.2.251 AP calculus exam concave ?

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Discussion Overview

The discussion revolves around the properties of the function \( g = e^{-t^2} \), specifically focusing on its concavity and inflection points as related to the AP calculus exam. Participants explore the implications of the first and second derivatives of the function within a specified interval.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant suggests that the function \( g = e^{-t^2} \) has an inflection point at \( x=1 \).
  • Another participant states that \( g' > 0 \) for all \( x \in (0,2) \), indicating that \( g \) is increasing.
  • A claim is made that \( g'' = e^{-x^2} > 0 \) for all \( x \in (0,2) \), suggesting that \( g \) is concave up.
  • Several participants express confusion or seek clarification regarding the reasoning behind the claims made about concavity and inflection points.
  • A participant presents the derivative \( g'(x) = \int_0^x e^{-t^2} \, dt \) as an area accumulation function, providing inequalities for \( g' \) at different points.

Areas of Agreement / Disagreement

There is no consensus on the existence of an inflection point at \( x=1 \), as participants express differing views on the concavity of the function and its implications.

Contextual Notes

Participants have not fully resolved the implications of the second derivative test and its relationship to the function's concavity and inflection points. The discussion includes varying interpretations of the behavior of the function within the specified interval.

karush
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View attachment 9441

ok this one baffled me a little
but isn't $g = e^{-t^2}$and the graph of that has an inflection point at x=1
 

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$g’ > 0 \text{ for all } x \in (0,2) \implies g \text{ is increasing}$

$g’’ = e^{-x^2} > 0 \text{ for all } x \in (0,2) \implies g \text{ is concave up}$
 

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karush said:
how?

how what?
 
$\displaystyle g'(x) = \int_0^x e^{-t^2} \, dt$ is the area accumulation function shown in the graph.

$$\color{red}{g'(0.5) = \int_0^{0.5} e^{-t^2} \, dt} < \color{blue}{g'(1) = \int_0^1 e^{-t^2} \, dt} < \color{green}{g'(1.5) = \int_0^{1.5} e^{-t^2} \, dt}$$
 

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