Question

Please explain this one for me
http://i.imgur.com/CPOHPAz.png
I don't know why g''(0)>0 => (0,0) is relative minimum

Answer 1

So you understand that the point is a critical point but not why it is the relative minimum?

Answer 2

yes

Answer 3

The second derivative is used to determine concavity so if it is greater than 0 it is concave up, meaning that the point must be a minimum

Answer 4

If it was concave down, or less than 0, it would be a relative maximum.

Answer 5

Ah I got it clearer now, thanks for your explaining

Answer 6

No problem :)

Answer 7

Remember, the relative minimum at critical points. Since we are dealing with a polynomial, our critical points will be where-ever the first derivative is 0. Setting g'(x) = 0 and solving for x suggests that g(x) = 0 when x = 0, 12/5, 6. So any of these x-values could have either a relative maximum or a relative minimum.

To figure out the type of extremum at the critical values, we find the second derivative of the function. Basically, the way it goes is: f''(x) > 0, function is concave up at that x, and so that value (x, f(x)) is a minimum. If f''(x) < 0, function is concave down at that x, so there is a maximum at that (x, f(x)).

That's basically what was done here. They found the second derivative g''(x) for g(x) and then checked each critical value to see if the g''(x) was greater than 0 or less than 0. If it's greater, then the function is concave up, meaning there is a relative minimum at that point whereas if it's less than 0, then function is concave down, so there will be a relative maximum.

@dangbeau