00:01
Here we have given an equation that is f of x equals to x square plus 3x plus a when x is less than equals to 1 and f x is b x plus 2 when x is greater than 1 and we have to find a and b.
00:23
So, we also given f x is continuous and differentiable.
00:33
So, for f x to be continuous f of x plus that is the right hand limit has to be equals to f of x and has to be equals to f of x minus that is left hand limit this is your right hand limit this is your left hand limit.
01:02
So, here we will solve it f of x plus this is given b x plus 2 also this has to be continuous at the broken point.
01:21
So, here the point is x equals to 1.
01:25
So, this has to be valid at x equals to 1.
01:31
So, this is right hand limit equals to at f equal at f x that is at x equals to 1 exact.
01:40
So, it is x square plus 3x plus a.
01:44
So, we will put x equals to 1 it will be b plus 2 equals to 1 plus 3 plus a here we got b minus a equals to 2 this is your first equation.
02:01
Now f x also has to be differentiable...