May 10, 2017

D.

#### Explanation:

$f$ is differentiable on each of $\left(0 , 2\right)$ and $\left(4 , 6\right)$.
Since differentiability implies continuity, $f$ is also continuous on each of the open intervals.

So, for any interval $\left[a , b\right]$ entirely within one of these open intervals, we can apply the Mean Value Theorem

$f$ is continuous on $\left[a , b\right]$ and
$f$ is differentiable on $\left(a , b\right)$

Therefore, there is a $c$ in $\left(a , b\right)$ such that

$f \left(b\right) - f \left(a\right) = f ' \left(c\right) \left(b - a\right)$.

Furthermore, since $f ' \left(c\right) = 1$, we see that

So, for any interval $\left[a , b\right]$ entirely within one of these open intervals,

$f \left(b\right) - f \left(a\right) = \left(b - a\right)$.

Therefore,

$f \left(5.5\right) - f \left(4.5\right) = 1 = f \left(1.5\right) - f \left(0.5\right)$

The function

$f \left(x\right) = \left\{\begin{matrix}x + 2 & \text{if" & 0 < x < 2 \\ x-2 & "if} & 4 < x < 6\end{matrix}\right.$ satisfies the given condition but fails A, B, and C. The graph is shown below.