Answer:
A)

B) Continuous but not differentiable.
Step-by-step explanation:
So we have the piecewise function:

A)
To write the differentiated piecewise function, let's differentiate each equation separately. Thus:
1)
![\frac{d}{dx}[\frac{1}{2}x+5}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7B1%7D%7B2%7Dx%2B5%7D%5D)
Expand:
![\frac{d}{dx}[\frac{1}{2}x]+\frac{d}{dx}[5]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7B1%7D%7B2%7Dx%5D%2B%5Cfrac%7Bd%7D%7Bdx%7D%5B5%5D)
The derivative of a linear equation is just the slope. The derivative of a constant is 0. Thus:
![\frac{d}{dx}[\frac{1}{2}x+5}]=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7B1%7D%7B2%7Dx%2B5%7D%5D%3D%5Cfrac%7B1%7D%7B2%7D)
2)
![\frac{d}{dx}[6]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B6%5D)
Again, the derivative of a constant is 0. Thus:
![\frac{d}{dx}[6]=0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5B6%5D%3D0)
3)
We have:
![\frac{d}{dx}[x+4]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%2B4%5D)
Expand:
![\frac{d}{dx}[x]+\frac{d}{dx}[4]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5D%2B%5Cfrac%7Bd%7D%7Bdx%7D%5B4%5D)
Simplify:

Now, let's substitute our original equations for the differentiated equations. The inequalities will stay the same. Therefore:

B)
For a function to be differentiable at a point, the function <em>must </em>be a) continuous at that point, and b) the left and right hand derivatives must be equivalent.
Let's first determine if the function is continuous at the point. Remember that a function is continuous at a point if and only if:

Let's find the left hand limit of f(x) at it approaches 2.

Since it's coming from the left, let's use the third equation:

Direct substitution:

So:

Now, let's find the right-hand limit:

Since we're coming from the right, let's use the first equation:

Direct substitution:

Multiply and add:

So, both the left and right hand limits are equivalent. Now, find the limit at x=2.
From the piecewise function, we can see that the value of f(2) is 6.
Therefore, the function is continuous at x=2.
Now, let's determine differentiability at x=2.
For a function to be differentiable at a point, both the right hand and left hand derivatives must be equivalent.
So, let's find the derivative of the function as x approahces 2 from the left and from the right.
From the differentiated piecewise function, we can see that as x approaches 2 from the left, the derivative is 1.
As x approaches 2 from the right, the derivative is 1/2.
Therefore, the right and left hand derivatives are <em>not</em> the same.
Thus, the function is continuous but <em>not</em> differentiable.