Two true/false questions I don't understand

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Homework Help Overview

The discussion revolves around understanding true/false statements related to calculus concepts, specifically antiderivatives and inflection points. Participants are exploring the implications of conditions on derivatives and their relationships to critical points and concavity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of statements regarding antiderivatives and critical points, particularly the relationship between the second derivative being zero and inflection points. There is also exploration of the conditions under which a critical point is defined.

Discussion Status

Some participants have offered insights into the nature of antiderivatives and critical points, while others are seeking clarification on the definition of inflection points and the role of the second derivative. Multiple interpretations of the conditions are being explored without reaching a consensus.

Contextual Notes

There is an ongoing discussion about the definitions and implications of critical points and inflection points, including the potential for undefined derivatives at critical points and the necessity of checking for sign changes in the second derivative.

IntegrateMe
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1. If a is a positive, then the function h(x) = (ln(ax^2)+x)/x is an antiderivative of j(x) = (2-ln(ax^2))/x^2

So, I used Wolfram and took the integral of j(x) with different values for a and always got ln(ax^2)/x, so I put false. However, the answer is true, and I can't figure out why!

2. If x = a is a critical point of a function m(x), then m'(a) = 0.

For this one I put true, and the answer is false. Is it because m'(a) can also be undefined?

Thank you!
 
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The difference between (ln(ax^2)+x)/x and ln(ax^2)/x is a constant. What constant? So they are both antiderivatives of (2-ln(ax^2))/x^2. And sure, m'(a) might be undefined at a critical point.
 
How about this:

If f"(a) = 0, then f has an inflection point at x = a.

The answer is false, but I thought inflection points were where the second derivative is equal to 0? Any clarification on this one?
 
IntegrateMe said:
How about this:

If f"(a) = 0, then f has an inflection point at x = a.

The answer is false, but I thought inflection points were where the second derivative is equal to 0? Any clarification on this one?

Same deal. An inflection point is where the concavity changes from concave up to concave down or vice versa. Define f(x)=x^2 for x>=0 and -x^2 for x<0. The derivative exists and is continuous and x=0 is an inflection point, but the second derivative doesn't exist there.
 
But this problem is explicitly saying that the second derivative is equal to 0 at a. So shouldn't a on the original graph be an inflection point?
 
IntegrateMe said:
But this problem is explicitly saying that the second derivative is equal to 0 at a. So shouldn't a on the original graph be an inflection point?

Yeah, I was going backwards. Think about f(x)=x^4. Is x=0 an inflection point?
 
No, it's not! I can see what you mean graphically, but can you explain it more clearly? I'm sorry I didn't follow the first time.
 
IntegrateMe said:
No, it's not! I can see what you mean graphically, but can you explain it more clearly? I'm sorry I didn't follow the first time.

I mean that if f(x)=x^4 then f''(x)=0. But f(x) is concave up everywhere.
 
So if we're given a function and asked to find inflection points, how can we verify that it's an inflection point besides checking that the second derivative is equal to 0? Do we have to check the first derivative and see if it changes from +/- or -/+ at that point as well?
 
  • #10
IntegrateMe said:
So if we're given a function and asked to find inflection points, how can we verify that it's an inflection point besides checking that the second derivative is equal to 0? Do we have to check the first derivative and see if it changes from +/- or -/+ at that point as well?

No, I think you have to check if the second derivative changes sign.
 

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