00:01
In this problem we are provided with the function g of teta which equals to 4 times teta minus 6 times sine of teta where teta belongs to closed interval 0 up to pi.
00:20
We are asked to find out the maximum and the minimum value of the function g of teta.
00:27
So for this let us first find out the critical points.
00:30
So differentiating the given function, we have 4 minus 6 times the derivative of sine of theta which is cost of theta.
00:39
Next, we equate this first derivative to 0 and we solve for theta.
00:45
So we get 4 minus 6 times cost of theta equals to 0 which implies that cost of theta equals to 4 over 6 which simplifies to theta equals to cos inverse of.
01:00
Of 2 over 3.
01:03
So now we must check the value of the function when teta equals to 0 pi and cos inverse of 2 over 3.
01:14
So let us begin with 0 we have g of 0 to be equal to 4 times 0 minus 6 times sine of 0 4 times 0 is 0 and we know that sign of 0 equals to 0.
01:29
So 6 times 0 will also be 0.
01:32
So we get the value of g of 0 to be 0.
01:34
Next we have g of pi.
01:37
So this equals to 4 times pi minus 6 times sine of pi.
01:43
Sign of pi equals to 0.
01:45
So we get g of pi to be 4 times pi which is approximately equal to 12 .5 double 6 4.
01:56
Next we evaluate g of cos inverse of 2 over 3.
02:04
So we have 4 times cost inverse of 2 over 3 minus 6 times sign of cost inverse of 2 over 3...