00:01
So here we're given our function and the point 2 comma 0.
00:04
So what we need to do is find the points on the graph of this function that are closest to this particular point by minimizing the square of the distance.
00:17
Okay, so let's go ahead and get started.
00:20
So now if a point is on our function f of x, it's going to have coordinates x comma y, but we know how to express the y in terms of x.
00:30
That's just the function.
00:32
So it's square root of x squared plus 1.
00:35
And now we're going to define the square distance function.
00:40
So that'll be d of x, and that is equal to x minus 2 squared plus the y value, which is square root of x squared plus 1, and it's minus 0 squared.
00:56
Okay? so let's go ahead and simplify that a little bit.
01:00
So the first one will leave it as x minus 2 squared, but the second one, we're squaring a square root, so that's just x squared plus one.
01:09
Now we're going to go ahead and take the derivative.
01:12
Okay, so d prime of x is equal to two times x minus two plus two x.
01:19
Simplify that a little bit.
01:21
So that's 2x plus 2x, which is 4x minus 4.
01:26
So now to find critical values, we set this equal to zero and solve for x...