00:03
Here, given f of x is equal to 2 plus 7x plus 63 upon x.
00:09
Name it equation 1.
00:12
Now we have to find out the local maximum and local minimum point.
00:19
So differentiating equation 1 with respect to x, we get f dash x is equal to 7 minus 63 upon x square.
00:33
Now in order to find out the local maximum or local minimum, using the formula f dash x is equal to 0, we get 7 minus 63 upon x square is equal to 0, which implies 7x square is equal to 63, which implies x square is equal to 9, which implies x is equal to plus minus 3.
01:03
So at x is equal to 3, we have to find out f double dash 3.
01:12
So again differentiating equation 1 twice with respect to x, we get f double dash x is equal to 63 multiplied by 2 upon x cube, which is equal to 126 upon x cube.
01:44
Therefore, f double dash 3 is 4 .7, which is greater than 0.
01:56
Hence, at x is equal to 3, local minimum is attained.
02:14
Now, the other value that is x is equal to minus 3, we have to evaluate f double dash minus 3...