Update meanValueTheorem3.tex

meanValueTheorem/exercises/meanValueTheorem3.tex

@@ -42,26 +42,19 @@
 
 $0<g'(x)\leq2$ for all $x$ in $(3,5)$.
 \begin{hint}
-Since the function $g$ satisfies the conditions of the Mean value theorem (MVT), there exists a points $c$ in $(3,5)$ such that $g'(c)=\answer{3}$.
-\end{hint}
-\begin{hint}
-Since  $g'(c)=\answer{3}$,\\
- it follows that $g'(c)>2$.
+Since the function $g$ satisfies the conditions of the Mean value theorem (MVT), there exists a points $c$ in $(3,5)$ such that $g'(c)=\answer{3}$. Thus, $g'(c)>2$.
 \end{hint}
+
 \begin{multipleChoice}
 \choice{True}
 \choice[correct]{False}
 \end{multipleChoice}
 
 $0<g(x)\leq 8$ for all $x$ in $(3,5)$.
 \begin{hint}
-Since $g'(x)>0$ on $(3,5)$, the function $g$ is increasing on the interval $(3,5)$. 
+Since $g'(x)>0$ on $(3,5)$, the function $g$ is increasing on the interval $(3,5)$. Thus,  $g(x)\le g(5)$, for all $x$ in $(3,5)$.
 \end{hint}
-\begin{hint}
-Since  the function $g$ is increasing (and continuous on $[3,5]$) on the interval $(3,5)$, it follows that
 
- $g(x)\le g(5)$, for all $x$ in $(3,5)$.
-\end{hint}
 \begin{multipleChoice}
 \choice[correct]{True}
 \choice{False}
@@ -70,4 +63,4 @@
 
 
 \end{exercise}
-\end{document}
+\end{document}