00:02
Let's investigate the mean value theorem as it applies to the function f of x equals x to the two -thirds on the interval from negative 1 to 8.
00:12
Well, let's go ahead and take the derivative.
00:20
Derivative of x to the 2 -thirds is 2 -thirds x to the 1 minus 2 -thirds, which is 2 -thirds x to the negative 1 -3, which i can write as 2 over 3 times the cube root of x.
00:48
Let's now find the slope of the second line between x equals negative 1 and x equals 8.
01:00
And so that's going to be 8 to the 2 3 minus negative 1 to the 2 thirds over 8 minus negative 1 to the 2 thirds over 8 minus negative 1.
01:19
Taking a number to the two -thirds power means you take the cube root and then you square it.
01:24
The cube root of 8 is 2.
01:27
2 squared is 4, minus the cube root of negative 1 is negative 1, squared is positive 1, over 8 minus negative 1.
01:48
4 minus 1 is 3, 8 minus negative 1 is 9, 8 minus negative 1 is 9, and so we get a slope of one -third.
01:59
And so now we want to find any values of c that when plugged into the derivative, give us a slope of one -third.
02:11
And so we're going to say 2 over 3 times a cube root of c equals 1 -third, multiply both sides by 3, and the 3s will cancel...