Given the function: g(x)= \left \{ {{x+b, x \ \textless \ 0 } \atop {cos(x), x \geq 0}} \right. Find the value of b, if any, that will make the function differentiable at x = 0.…

Question

Given the function:


`g(x)= \left \{ {{x+b, x \ \textless \ 0 } \atop {cos(x), x \geq 0}} \right.`

Find the value of b, if any, that will make the function differentiable at x = 0.

A. 0
B. 1
C. 2
D. No such value exists
E. There is not enough information to find the value

Can someone confirm that the answer is D, no such value exists? I don't think there's a value where the function could be continuous and have the same slope on each piece.

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

--------------------------------

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.