# Thread: inflection point, concave up and down question

1. ## inflection point, concave up and down question

f(x)=x^3-3x^2+3
1.First derivative 3x^2-6x
2. intervals where f is increasing. 3x^2-6x=0, x=0, x=2 Function is increasing from -infinity to 0 and from 2 to infinity
3. decreasing (0,2)
4. local max point (0,3)
5. local min point (2,-1)
6. second derivative 6x-6
I hope everything until now is correct
7. inflection point 6x-6=0, x=1 Is it correct to say that at x=1 function changes from concave down to concave up?
8. what are the intervals where function is concave up and concave down?
Thank you very much

2. Originally Posted by oceanmd
f(x)=x^3-3x^2+3
1.First derivative 3x^2-6x
2. intervals where f is increasing. 3x^2-6x=0, x=0, x=2 Function is increasing from -infinity to 0 and from 2 to infinity
3. decreasing (0,2)
4. local max point (0,3)
5. local min point (2,-1)
6. second derivative 6x-6
I hope everything until now is correct
7. inflection point 6x-6=0, x=1 Is it correct to say that at x=1 function changes from concave down to concave up?
8. what are the intervals where function is concave up and concave down?
Thank you very much
All your work is correct! For #8, you already said it in #7. The concavity of f(x) changes at x=1, so your intervals would be concave down $(-\infty, 1)$ and concave up on $(1, \infty)$

3. ## when the second derivative is a positive number

Please explain if I am wrong. Thank you very very much.
Inflection point
F(x) = 3x^2 + 12x + 3
Please correct me if I am wrong
1. We need to find first derivative to find minimum and maximum points
First derivative = 6x +12, 6x + 12 = 0, x = -2
2.Local minimum is (-2, -9)
3. Second derivative = 6 What does it mean? Where is the inflection point? There is no inflection point. Right? Is it correct to say that because the second derivative is a positive number, the graph is always concave up. For this particular function the graph is concave up from negative infinity to positive infinity.

4. Originally Posted by oceanmd
Please explain if I am wrong. Thank you very very much.
Inflection point
F(x) = 3x^2 + 12x + 3
Please correct me if I am wrong
1. We need to find first derivative to find minimum and maximum points
First derivative = 6x +12, 6x + 12 = 0, x = -2
2.Local minimum is (-2, -9)
3. Second derivative = 6 What does it mean? Where is the inflection point? There is no inflection point. Right? Is it correct to say that because the second derivative is a positive number, the graph is always concave up. For this particular function the graph is concave up from negative infinity to positive infinity.
Is it correct to say that because the second derivative is a positive number, the graph is always concave up? Yes.

**Always make a new thread for a new problem, okay?

5. mollymcf2009,
Thank you very much.
Is it correct to say that there is no inflection point?

6. Originally Posted by oceanmd
mollymcf2009,
Thank you very much.
Is it correct to say that there is no inflection point?
Yes