Piecewise Functions - 2

We see that both $x^3 + 2x^2 + x + c$ and $e^x$ are differentiable in their domain. So, we only need to make the function differentiable at $x = b$ .

Since $f$ is differentiable at $x = b$ ,

$\text{Left hand differentiation = Right hand differentiation (at } x = b)$

$3b^2 + 4b + 1 = e^b$

$b$ is an integer, so LHS of this equation will always be an integer. However, RHS will be an integer only if $b = 0$ .

If $b = 0$ , LHS = RHS, so $b = 0$ is the solution to this equation.

Since $f$ is differentiable at $x = b$ , it is implied that it is also continuous at $x =b$ .

$\displaystyle\lim_{x\rightarrow b^{-} } f(x) = f(b)= \displaystyle\lim_{x \rightarrow b^{+}} f(x)$

$b^3 + 2b^2 + b + c = e^b .$

When $b = 0$ , we get $c = e^0 = 1$

So, the answer is $0 + 1 = \boxed{1}$