00:01
Function f of x is equal to 3 minus 2x minus 5x squared on the close interval from negative 3 to 3.
00:08
So we first go ahead and take the derivative of our function.
00:11
So we find f prime of x.
00:13
So you take the derivative here term by term and get the derivative is equal to, well, negative 2 minus 10x or negative 10x minus 2 either way.
00:24
Okay.
00:25
And now, find the critical values, we just take our derivative and we set it equal.
00:30
To 0.
00:31
And then solve for x.
00:32
So we'd have negative 2 minus 10x is equal to 0.
00:36
That implies that negative 10x is equal to 2.
00:40
So we have that x is equal to well negative 2 over 5 or negative 1 5th.
00:45
So our critical value, right, where the derivative is equal to 0 happens at x being equal to negative 1 5th.
00:54
Okay.
00:57
So then we just list the critical values.
01:00
And the endpoints.
01:02
So our two endpoints are negative three and three.
01:04
So we have negative three.
01:05
Then we have our critical value of negative one -fifth, and we have three.
01:09
Then we evaluate the function at these three points...