00:01
In the problem, the given function fx, that is cos x, for the values x less than 0 and ax plus b for the values x greater than equal to 0, it is given that f will be differentiable everywhere.
00:18
Now, in that case, if f is differentiable, then at the point 0, the f will be differentiable and the left hand side differentiation and right hand side differentiation, both values will be similar.
00:29
Therefore, the left hand side differentiation for the given function fx, that will be cos x and f prime 0 minus, that will be for cos x, we will get minus sin x and minus sin 0, that will give us 0.
00:45
If we take the right hand side, then in that case, we will get f prime 0 plus, we will take the function ax plus b for x greater than equal to 0, differentiation of ax plus b, that will give us x.
01:00
So, these two values have to be same for f to be differentiable everywhere.
01:05
So, this is giving us a equal to 0...