00:01
In your question, we're asked to determine the intervals where the graph of the function is concave up and concave down.
00:06
So to explore concavity, we need to take our second derivative.
00:11
We need to look at our second derivative.
00:13
So to get there, we have to get our first derivative.
00:16
The 4 thirds would come down and multiply the 3, giving us 8 thirds.
00:21
X, and the power drops by 1, taking it to a 1 third power.
00:26
Power.
00:28
Plus, the 1 third would come down and multiply the 3, making it a 1x to the negative 2 thirds power when we subtract 1 from our power.
00:41
Now we'll go ahead and take our second derivative, and the 1 third would come down and multiply our front number, giving us 8 ninths x to the negative 2 thirds power.
01:01
And in the next part, negative 2 thirds would would multiply our 1, making it a minus 2 thirds, and that would be x to the negative 5 thirds power.
01:19
Now we can work to solve this.
01:23
We want to set it equal to 0.
01:27
What we're looking for is places where the second derivative equals 0, or the second derivative doesn't exist, and those are places where we could possibly change concavity.
01:39
So i think what what i want to do here is add 2 thirds x to the negative 5 thirds to both sides, giving me 8 ninths x.
01:50
I'm going to move this down here to positive 2 thirds in the denominator, equals a positive 2 thirds with an x to the 5 thirds in the denominator.
02:04
I would then cross multiply, supply, giving me 24x to the 5 thirds power equals 18x to the 2 thirds power.
02:20
Now we'll subtract and bring both x terms back to the left side, set it equal to 0, and i'll now factor out a 6 in the least power, which is x to the 2 thirds, giving me 4x minus 3 equals 0.
02:49
We can then solve.
02:51
We would have 6x to the 2 thirds equals 0.
02:55
That's just x equals 0...