Question

Given the function:

Answer 1

You are correct. The answer is choice D

The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0

To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b. 

So the piecewise function becomes
$g(x) = \begin{cases}x+1 \ \text{ if } \ x \ < \ 0\\ \cos(x) \text{ if } \ x \ge 0\end{cases}$
after plugging in b = 1

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Now differentiate each piece with respect to x to get
$g'(x) = \begin{cases}1 \ \text{ if } \ x \ < \ 0\\ -\sin(x) \text{ if } \ x \ge 0\end{cases}$
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0

So there's this disconnect on g ' (x) meaning its not continuous

Therefore, the value b = 1 will not work. 

So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.