00:01
Okay, so let's get started by recalling the definition of local minimum.
00:07
Well, we know that g has a local minimum at x equal to c if and only if there exists an open interval i, so open interval containing c such that f of x is greater than or equal to f of c for every x belonging to i.
00:40
Perfect.
00:41
Now, we claim that if g has a local minimum at x equal to c, then g is going to have a local maximum at x equal to negative c.
00:55
So, g has a local max at x equal to negative c.
01:05
Okay, how can we prove this? well, pretty easy.
01:10
Let me consider the open interval j, which is going to be negative i as above.
01:18
That is, this one is the interval consisting of points negative x such that x belongs to i.
01:29
Okay, let me draw a figure.
01:33
Well, if this one is c, here we have i, here we have negative c, then j is going to be, oh, here we have zero.
01:48
Perfect.
01:49
Okay, then j is going to be this, this open interval.
01:54
Perfect.
01:55
Okay, now let's observe that we know that f and that g is odd...