00:01
In this question, we are asked to find the global maximum and minimum points of the function f.
00:06
And to do that, we first need to solve the equation f prime of x equals to 0, where f prime of x is the derivative of the function f, and in our case, f prime of x equals to 2 times 3x squared minus 3 times 2x minus 12, which simplifies to 6x squared minus 6x minus 12.
00:33
And that gives us a quadratic equation for finding x.
00:40
We can divide everything by 6.
00:42
We are going to get x squared minus x, minus 2 equals 0.
00:48
And the discriminant of this equation is negative 1 squared minus 4 times negative 2.
00:57
And that's going to be 9.
00:59
Therefore, the square root of the discriminant is 3.
01:03
And the roots of the equation are x equals 1 plus minus 3 over 2.
01:08
So one solution is 4 over 2 which is 2 and the other solution is negative 2 over 2 which is negative 1.
01:18
So these are the so -called stationary points of the function.
01:24
And now to find the global maximum and minimum values, we need to calculate the value of the function f at the stationary points.
01:37
And also we need to calculate the value of f at the end points of the interval.
01:43
Negative 3 and 3 and then we need to find the largest and the smallest numbers so for convenience i'm going to bring down the function f all right so f of 2 equals to 2 times 8 minus 3 times 4 minus 24 plus 5 3 times or 2 times 8 is 16 16 plus 1 20 is negative 3, negative 3 minus 12 is negative 15.
02:38
Now, f of negative 1.
02:40
F of negative 1 equals to negative 2 minus 3 plus 12 plus 5.
02:48
And that's going to be 12...