1. ## Question About Pt of Inflection

Does a point of inflection occur if the second derivative graph does not have a derivative at the point, but the first derivative has a max/min at the point?

2. Originally Posted by WhoCares357
Does a point of inflection occur if the second derivative graph does not have a derivative at the point, but the first derivative has a max/min at the point?
What do you mean "does not have a derivative at the point"? Do you mean undefined, constant, zero?

In terms of f'(x), a point of inflection occurs where f'(x) has a max/min, like you said. To solve for this you take the derivative of f'(x), which is f''(x) and set it equal to zero. So if you are looking at the f''(x) graph then you need to be looking for when f''(a)=0.

3. Originally Posted by Jameson
What do you mean "does not have a derivative at the point"? Do you mean undefined, constant, zero?

In terms of f'(x), a point of inflection occurs where f'(x) has a max/min, like you said. To solve for this you take the derivative of f'(x), which is f''(x) and set it equal to zero. So if you are looking at the f''(x) graph then you need to be looking for when f''(a)=0.
Well, the limit for f'' at that point is dne, because the limit is different from both sides. It jumps over the x-axis. Is that still a point of inflection?

4. Originally Posted by WhoCares357
Well, the limit for f'' at that point is dne, because the limit is different from both sides. It jumps over the x-axis. Is that still a point of inflection?
How about posting the problem so I can use more concrete terms?

5. Originally Posted by Jameson
How about posting the problem so I can use more concrete terms?
4 b

6. Ok, thank you. Makes more sense now. So you are correct that the second derivative doesn't exist at x=0, but this isn't a requirement for an inflection point. If a is an inflection point then f'(a) exists and f''(a) changes sign at x=a. So if you show that the sign of f''(x) changes at x=0 and f'(0) exists you have an inflection point. You can use the graph to justify that before x=0, the f'(x) graph has a negative slope while after x=0 it has a positive one. Sometimes the AP tests get picky about topics and want you to answer a certain way so I'd check with your teacher or look up the answer when you can to make sure that they agree with this reasoning.

7. Originally Posted by Jameson
Ok, thank you. Makes more sense now. So you are correct that the second derivative doesn't exist at x=0, but this isn't a requirement for an inflection point. If a, is an inflection point then f'(a) exists and f''(a) changes sign at x=c. So if you show that the sign of f''(x) changes at x=0 and f'(0) exists you have an inflection point. You can use the graph to justify that before x=0, the f'(x) graph has a negative slope while after x=0 it has a positive one. Sometimes the AP tests get picky about topics and want you to answer a certain way so I'd check with your teacher or look up the answer when you can to make sure that they agree with this reasoning.
Thanks. Exactly what I needed.

8. Originally Posted by WhoCares357
4 b

f'(0) does not exist. No tangent can be drawn at x = 0. There is no point of inflection at x = 0.

9. Originally Posted by mr fantastic
f'(0) does not exist. No tangent can be drawn at x = 0. There is no point of inflection at x = 0.
How do can you be certain that f(0) does not exist?

10. Originally Posted by WhoCares357
How do can you be certain that f(0) does not exist?
I said f'(0) does not exist. The gradient of the tangent to f(x) is different on either side of x = 0. The point at x = 0 is a salient point.

11. Originally Posted by mr fantastic
I said f'(0) does not exist. The gradient of the tangent to f(x) is different on either side of x = 0. The point at x = 0 is a salient point.
But f'(0) does exist. This is the graph of f' and there is no hole or jump or asymptote at the point. Could the original not have been a peace wise function with the same slope at f(0)?

PS Sorry for the late response. It's been a busy week[end].

12. Originally Posted by WhoCares357
But f'(0) does exist. This is the graph of f' and there is no hole or jump or asymptote at the point. Could the original not have been a peace wise function with the same slope at f(0)?

PS Sorry for the late response. It's been a busy week[end].
Sorry, I completely overlooked the bit that said that it was a graph of f'(x).