Find the values of `a` and `b` so that the function `f(x)={x^2+3x+a ,\ \ if\ xlt=1b x+2,\ \ \ if\ x >1` is differentiable at each `x in R` .

Explanation

Given that,
`f(x)=x^2+3x+a `,for `x le1`
`=bx+2`, for `x gt1`
The function is everywhere differentiable.
So, it is also continuous.
Therefore,
`lim_(x->1)f(x) =lim_(x->1^+) (f(x)`
`=lim_(x->1)x^2+3x+a =lim_(x->1) bx+2`
`=>4+a=b+2`
`=>a-b=-2.....(1)`
Now, f(x) is differentiable at x=1, therefore,
`lim_(x->1^-)(f(x)-f(1))/(x-1) =lim_(x->1^-+)(f(x)-f(1))/(x-1)`
`=>lim_(x->1)((x^2+3x+a)-(a+4))/(x-1)=lim_(x->1)((bx+2)-(b+2))/(x-1)`
`=>lim_(x->1)(x^2+3x-4)/(x-1)=lim_(x->1)(b(x-1))/(x-1)`
`=>lim_(x->1)((x+4)(x-1))/(x-1)=lim_(x->1)(b(x-1))/(x-1)`
`=>lim_(x->1) (x+4) =lim_(x->1)b`
`=>b=5`
so, `a=3`