Fix typos

computationsForGraphingFunctions/digInComputationsForGraphingFunctions.tex

@@ -198,7 +198,7 @@
 This is only a \textbf{possible} inflection point; we still have to check whether concavity changes there.\\
 
  We have that $f''(x)<0$ for $x<\frac{1}{2}$, therefore
-$f$ is concave \wordChoice{\choice{up} \choice[correct]{down}}on $\left(-\infty,\frac{1}{2}\right)$.\\
+$f$ is concave \wordChoice{\choice{up} \choice[correct]{down}} on $\left(-\infty,\frac{1}{2}\right)$.\\
 
 Similarly, $f''(x)>0$ for $x>\frac{1}{2}$, therefore
 $f$ is concave \wordChoice{\choice[correct]{up} \choice{down}} on $\left(\frac{1}{2},\infty\right)$.
@@ -475,12 +475,12 @@
 On $(-\infty, 0)$, $f''$ has one zero, namely $x =
 \answer[given]{-2}$.  The sign of $f''$ changes from
 \wordChoice{\choice{positive to negative} \choice[correct]{negative to
-    positive}]} at this point.
+    positive}} at this point.
 
       On $(0, \infty)$, $f''$ has one
         zero, namely $x = \answer[given]{\frac{1}{\sqrt{6}}}$. The
         sign of $f''$ changes from \wordChoice{\choice{positive to
-            negative} \choice[correct]{negative to positive}]} through
+            negative} \choice[correct]{negative to positive}} through
           this point.
 
 \begin{image}