If $f(x) = x, x ≤ 1$ and $f(x) = x^2+bx+c,x>1$ and $f'(x)$ exists finitely for all $x∈R$, then

$b=-1,c=1$

f(x) is differentiable at x = 1

$∴\underset{h→0}{\lim}\frac{f(1+h)-f(1)}{h}=\underset{h→0}{\lim}\frac{f(1-h)-f(1)}{-h}$

Now, $\underset{h→0}{\lim}\frac{f(1+h)-f(1)}{h}=\underset{h→0}{\lim}\frac{(1+h)^2+b(1+h)+c-1}{h}$

$=\underset{h→0}{\lim}\frac{h^2+(2+b)h+b+c}{h}$

$\underset{h→0}{\lim}\frac{f(1-h)-f(1)}{-h}=\underset{h→0}{\lim}\frac{1-h-1}{-h}=1$

The two limits can be equal if 2 + b =1, b + c = 0