Find the values of `aa n db` so that the function `f(x)={x^2+3x+a ,ifxlt=1b x+2,ifx >1` is differentiable at each `x in Rdot`

Answer

N/a

Explanation

We have, `f(x)={{:(x^(2)+3x+p, ifxle1),(qx+2,ifx gt1):}` is differentiable at `x = 1`
`:. Lf'(1)=underset(xrarr1^(-))(lim)(f(x)-f(1))/(xrarr1)`
`=underset(xrarr1^(-))(lim)((x^(2)+3x+p)-(1+3+p))/(x-1)`
`=underset(hrarr0)(lim)([(1-h)^(2)+
3(1-h)+p]-[1+3+p])/((1-h)-1)`
`= underset(hrarr0)(lim)([1+h^(2)-2h+3-3h+p]-[4+p])/(-h)`
`=underset(hrarr0)(lim)([h^(2)-5h+p+4-4-p])/(-h)=underset(hrarr0)(lim)(h[h-5])/(-h)`
`= underset(hrarr0)(lim) -[h-5] = 5`
`Rf'(1) = underset(x+1^(+))(lim)(f(x)-f(1))/(x-1)=underset(xrarr1^(+))(lim)((qx+2)-(1+3+p))/(x-1)`
`= underset(hrarr0)(lim)([q(1+h)+2]-(4+p))/(1+2-1)`
`= underset(hrarr0)(lim)([q+qh+2-4-p])/(h)=underset(hrarr0)(lim)(qh+(q-2-p))/(h)`
`rArr q - 2 - p = 0 rArr p-q=-2"......"(i)`
`rArr underset(hrarr0)lim(qh+0)/(h)=q`, [for existing the limit]
If `Lf'(1) = Rf'(1)`, then `5 = q`
`rArr p-5=-2rArr p = 5`
`:. p = 3` and `q = 5`