IVT for $a(x) = x+2+\frac{|x-2|}{x-2}$ If we have the function $a(x) = x+2 + \frac{|x-2|}{x-2}$, where $x$ is a member of $[1,5]$ and $y$ is a member of $[2,8]$, when we apply the Intermediate Value Theorem, why could it be either true or false? 
 A: $a(x) = 2 + x + \frac{|x-2|}{x-2}$ is not continuous on $[1,5]$ because it is not defined at $x=2$.  In fact $\lim_{x \to 2^-}a(x)=3, \lim_{x \to 2^+}a(x)=5$.  You therefore do not have the hypotheses of the theorem.
A: Notice that $\dfrac{|x-2|}{x-2}$ is equal to $-1$ if $x<2$ and equal to $1$ if $x>2$, so there's a jump discontinuity, and jump discontinuities are exactly what the intermediate value theorem says will not be there, if the function is continuous.
A: The function can be written piecewise as
$$
   f(x) = \begin{cases} x+1 & \text{if $1 ≤ x < 2$} \\
                        x+3 & \text{if $2 < x ≤ 5$} \end{cases}
$$
Notice that $f(2)$ is not defined.
The graph of $f$ is the line segment from $(1,2)$ to $(2,3)$ (not including the latter point), then the line segment from $(2,5)$ to $(5,8)$ (not including the former point).  You might want to draw it for reference.
We have
\begin{align*}
    \lim_{x\to 2^-} f(x) &= 3 \\
    \lim_{x\to 2^+} f(x) &= 5 
\end{align*}
So $\lim_{x\to 2} f(x)$ does not exist.  
The IVT requires that $f$ be continuous on a closed, bounded interval.  Right now $f$ is defined only on $[1,2)\cup(2,5]$, which is not a single interval (nor is it closed).  What's worse, even if we choose a value for $f(2)$, the fact that $\lim_{x\to 2}f(x)$ does not exist says that $f$ cannot be continuous at $2$.  So no matter what, we cannot satisfy the hypotheses of the IVT.
You say “why could it be either true or false?”  I am guessing what you mean is that why should the statement “there exists $x$ in $[1,5]$ such that $f(x) = c$” be true for some values of $c$ in $[2,8]$ and false for other values of $c$?  Let's verify it.  Is there an $x$ such that $f(x) = 6$?  Yes, $f(3) = 3+3 = 6$.
Is there an $x$ such that $f(x) = 4$? No, because $x$ is either less than 2 or greater than 2.  If $x < 2$, $f(x) = 4 \implies x+1 = 4 \implies x=3$, which is a contradiction.  If $x>2$, $f(x) = 4 \implies x+3 = 4 \implies x=1$, also a contradiction.  (This is easier to see on a graph; sorry I don't have the time to produce one.)
So there are some values of $c$ which have a corresponding $x$, but some values of $c$ which do not.  The reason this is consistent with the IVT, and not contradicting, it is that $f$ is not continuous.
