00:02
Okay, what we want to talk about today is we are given a function, f of x is equal to x squared plus a over x.
00:13
And the first thing we want to determine is we want to find a such that f of x has a local or relative minimum at x equal.
00:36
To 2.
00:38
Okay, so the first thing we need to do is take the derivative, so f prime of x is equal to 2x minus a over x squared.
00:51
And now what we're going to do is i'm actually going to put it over a common denominator.
00:57
And so this will be 2x cubed minus a over x squared.
01:05
And now we're going to set that derivative equal to zero.
01:10
So that will be a critical number and where possibly local minimums or local maximums will occur.
01:19
And so we're going to set the derivative equal to zero, which really means that numerator equal to zero.
01:28
And we know that they're saying we have a local minimum at two.
01:33
And so 0 is equal to 2 times 2 cubed minus a...