From f523fb8833a42b8cb0628de50e7c8013a8c7f9f9 Mon Sep 17 00:00:00 2001
From: pfatheddin <156558883+pfatheddin@users.noreply.github.com>
Date: Sun, 24 Mar 2024 18:10:01 -0400
Subject: [PATCH] Update meanValueTheorem3.tex

https://ximera.osu.edu/mooculus/meanValueTheorem/exercises/exerciseList/meanValueTheorem/exercises/meanValueTheorem3

Made two of the hints shorter by connecting the sentences and not repeating them.
---
 meanValueTheorem/exercises/meanValueTheorem3.tex | 15 ++++-----------
 1 file changed, 4 insertions(+), 11 deletions(-)

diff --git a/meanValueTheorem/exercises/meanValueTheorem3.tex b/meanValueTheorem/exercises/meanValueTheorem3.tex
index 704b0cfe6..160c0f9dd 100644
--- a/meanValueTheorem/exercises/meanValueTheorem3.tex
+++ b/meanValueTheorem/exercises/meanValueTheorem3.tex
@@ -42,12 +42,9 @@
 
 $0<g'(x)\leq2$ 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}
@@ -55,13 +52,9 @@
 
 $0<g(x)\leq 8$ 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}
\ No newline at end of file
+\end{document}