Graph of Integral: Find x-coords of Points of Inflection

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Homework Help Overview

The problem involves finding the x-coordinates of points of inflection for the graph of a function g on the interval (-2, 5). Participants are tasked with justifying their answers regarding the locations of these points.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss their reasoning for identifying x=0 and x=3 as points of inflection, with one participant questioning the justification provided by their tutor and teacher. There is exploration of the relationship between the function g and its derivatives, as well as the conditions under which inflection points occur.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on the conditions for inflection points and the implications of the derivatives. Some guidance has been offered regarding the relationship between g and its derivatives, but no consensus has been reached on the justification for the identified points of inflection.

Contextual Notes

Participants are navigating differing opinions from their tutor and teacher regarding the points of inflection, and there is mention of the need for a mathematical justification that may not align with their current explanations. The discussion also raises questions about the differentiability of g' and the existence of g''.

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Homework Statement


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Find the x-coordinate of each point of inflection of the graph of g on the open interval (-2, 5). Justify your answer

Homework Equations





The Attempt at a Solution


The answer (I think) is x=0 and x=3. Well, the thing is that I asked my tutor how to justify it and he said that it is only x=0 and not 3. But when I asked my teacher for the answer he checked and said it x=0, 3, but he couldn't tell me how to justify it. So now I'm wondering if I'm right or my tutor is.

The reason why I thought it was x=0 and x=3 is because from -2, the area as you move towards 0 is increasing at a greater and greater rate, therefore it's concave up. Then after 0 it's still increasing, but its rate of increase is decreasing, therefore concave down. Then at 2 it starts decreasing but the magnitude of the rate of decrease is still increasing (therefor still concave down) until x=3. Then after 3 it's still decreasing, but the magnitude of the rate is decreasing so it's concave up again. Then at 4 it starts to increase with increasing rate, and so g is still concave up.

But I'm pretty sure that on an exam they want a quick explanation in mathematical terms, not a whole long paragraph like that. My tutor said that because at x=3 there's a cusp so that's not a solution but I don't get why.:confused:
 
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Your answers are correct - the key lies in understanding two things:

When do inflection points occur in a function?

How are the functions g and f related?

The answer to the first question (it has something to do with derivatives...) will help you answer the second one (along with the 2nd Fundamental Theorem).

Sorry to be so vague, but I want you to figure it out for yourself.
 
Inflection points should occur where g''(x)=0...
And g'(x)=f(x), so g''(x)=f'(x), so they should occur where the slope changes from positive to negative or vice versa?
 
What's your justification that g'(x) is differentiable? Remember, inflection points can happen where g''(x) doesn't exist.
 

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