Help in sketching a CT signal with discontinuities
You are right, the solution manual is wrong and here is a piecewise linear definition of the signal $[x(t)+x(-t)]u(t)$ based on the acceptance of $\text{u}(0) = 0.5$ (i.e. the value at a discontinuity is the half way between right and left ends)
$$[x(t)+x(-t)]u(t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  t &lt; 0}}\\&amp;[1.5] &amp;\scriptstyle{\text{, t = 0}}\\&amp;3 &amp;\scriptstyle{\text{, 0&lt;t&lt;1}}\\&amp;[2] &amp;\scriptstyle{\text{, t = 1}}\\-&amp;2t+3 &amp;\scriptstyle{\text{,  1&lt;t&lt;2}}\\&amp;[-0.5] &amp;\scriptstyle{\text{, t=2}}\\&amp;0 &amp;\scriptstyle{\text{,  t &gt; 2}}\\
\end{align}
\end{cases}$$

Where the values within $[.]$ are points of discontinuities.

Let me put the steps to reach this conclusion in more details:
Looking at the provided graph of $x(t)$ we can deduce a piecewise definition of the signal. However we must be causious about the possible discontinuities, for the sake of mathematical (continuous) function theory, those isolated points of discontinuities can be omitted entirely or as an engineering convention the value of the signal can be defined to be the half of the sum of its values at the right and left limits of the function at the discontinuity point. I'll accept this second definition, as DSP is concerned more with engineering than with pure mathematics.

Therefore, we have the definition of $x(t)$ as follows:
$$x(t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  t &lt; -2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{, t = -2}}\\
&amp;t+1 &amp;\scriptstyle{\text{, -2&lt;t&lt;-1}}\\
&amp;[0.5] &amp;\scriptstyle{\text{, t = -1}}\\
&amp;1 &amp;\scriptstyle{\text{,  -1&lt;t&lt;0}}\\
&amp;[1.5] &amp;\scriptstyle{\text{, t=0}}\\
&amp;2 &amp;\scriptstyle{\text{,  0&lt;t&lt;1}}\\
&amp;[1.5] &amp;\scriptstyle{\text{,  t=1}}\\
&amp;-t+2 &amp;\scriptstyle{\text{,  1&lt;t&lt;2}}\\
&amp;0 &amp;\scriptstyle{\text{,  t&gt;2}}\\
\end{align}
\end{cases}$$

From which we can also deduce $x(-t)$ as follows:
$$x(-t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  -t &lt; -2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{, -t = -2}}\\
&amp;-t+1 &amp;\scriptstyle{\text{, -2&lt;-t&lt;-1}}\\
&amp;[0.5] &amp;\scriptstyle{\text{, -t = -1}}\\
&amp;1 &amp;\scriptstyle{\text{,  -1&lt;-t&lt;0}}\\
&amp;[1.5] &amp;\scriptstyle{\text{, -t=0}}\\
&amp;2 &amp;\scriptstyle{\text{,  0&lt;-t&lt;1}}\\
&amp;[1.5] &amp;\scriptstyle{\text{,  -t=1}}\\
&amp;t+2 &amp;\scriptstyle{\text{,  1&lt;-t&lt;2}}\\
&amp;0 &amp;\scriptstyle{\text{,  -t&gt;2}}\\
\end{align}
\end{cases}$$

Observe carefully how every occurence of $t$ is replaced with $-t$ both in the defining formula of the function and also in the definition of its intervals. We shall reformat intervals of $x(-t)$ with positive $t$ as follows:

$$x(-t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  t &gt; 2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{, t = 2}}\\
&amp;-t+1 &amp;\scriptstyle{\text{, 1&lt;t&lt;2}}\\
&amp;[0.5] &amp;\scriptstyle{\text{, t = 1}}\\
&amp;1 &amp;\scriptstyle{\text{,  0&lt;t&lt;1}}\\
&amp;[1.5] &amp;\scriptstyle{\text{, t=0}}\\
&amp;2 &amp;\scriptstyle{\text{,  -1&lt;t&lt;0}}\\
&amp;[1.5] &amp;\scriptstyle{\text{,  t=-1}}\\
&amp;t+2 &amp;\scriptstyle{\text{,  -2&lt;t&lt;-1}}\\
&amp;0 &amp;\scriptstyle{\text{,  t&lt;-2}}\\
\end{align}
\end{cases}$$

Now adjust the order of intervals of $x(-t)$ so that they are in left to right order:

$$x(-t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  t&lt;-2}}\\
&amp;t+2 &amp;\scriptstyle{\text{,  -2&lt;t&lt;-1}}\\
&amp;[1.5] &amp;\scriptstyle{\text{,  t=-1}}\\
&amp;2 &amp;\scriptstyle{\text{,  -1&lt;t&lt;0}}\\
&amp;[1.5] &amp;\scriptstyle{\text{, t=0}}\\
&amp;1 &amp;\scriptstyle{\text{,  0&lt;t&lt;1}}\\
&amp;[0.5] &amp;\scriptstyle{\text{, t = 1}}\\
&amp;-t+1 &amp;\scriptstyle{\text{, 1&lt;t&lt;2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{, t = 2}}\\
&amp;0 &amp;\scriptstyle{\text{,  t &gt; 2}}\\
\end{align}
\end{cases}$$

Now we can add these two signals for each respective interval. In fact we are here a bit lucky that the signals do have matching intervals, which would not be the case if the original signal $x(t)$ involved nonsymetric pieces in the first place. The addition produces the following signal:

$$x(t)+x(-t) = 
\begin{cases}
\begin{align} 
&amp;0 &amp;\scriptstyle{\text{,  t&lt;-2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{,  t=-2}}\\
&amp;(t+1)+(t+2)=2t+3 &amp;\scriptstyle{\text{,  -2&lt;t&lt;-1}}\\
&amp;[0.5]+[1.5]=[2] &amp;\scriptstyle{\text{,  t=-1}}\\
&amp;1+2=3 &amp;\scriptstyle{\text{,  -1&lt;t&lt;0}}\\
&amp;[1.5]+[1.5]=[3] &amp;\scriptstyle{\text{, t=0}}\\
&amp;2+1=3 &amp;\scriptstyle{\text{,  0&lt;t&lt;1}}\\
&amp;[1.5]+[0.5]=[2] &amp;\scriptstyle{\text{, t = 1}}\\
&amp;(-t+1)+(-t+1)=-2t+3 &amp;\scriptstyle{\text{, 1&lt;t&lt;2}}\\
&amp;[-0.5] &amp;\scriptstyle{\text{, t = 2}}\\
&amp;0 &amp;\scriptstyle{\text{,  t &gt; 2}}\\
\end{align}
\end{cases}$$

Finally we shall multiply this with the funciton $\text{u}(t)$ whose piecewise definition is:
$$\text{u}(t) =
\begin{cases}
\begin{align}
&amp;0 &amp;\scriptstyle{\text{, t&lt;0}}\\
&amp;[0.5] &amp;\scriptstyle{\text{, t=0}}\\
&amp;1 &amp;\scriptstyle{\text{, t&gt;0}}\\
\end{align}
\end{cases}$$

And the resulting function $[x(t)+x(-t)]\text{u}(t)$ was already provided in the first part above...