From acba914642f95153747544cc7726e52155b68983 Mon Sep 17 00:00:00 2001
From: pfatheddin <156558883+pfatheddin@users.noreply.github.com>
Date: Sun, 24 Mar 2024 19:17:42 -0400
Subject: [PATCH] Update LinearApproximations4.tex

https://ximera.osu.edu/mooculus/linearApproximation/exercises/exerciseList/linearApproximation/exercises/LinearApproximations4

They should be able to do this without any hint. I shortened the hint.
---
 .../exercises/LinearApproximations4.tex              | 12 ++++--------
 1 file changed, 4 insertions(+), 8 deletions(-)

diff --git a/linearApproximation/exercises/LinearApproximations4.tex b/linearApproximation/exercises/LinearApproximations4.tex
index cae679aff..adb6c431e 100644
--- a/linearApproximation/exercises/LinearApproximations4.tex
+++ b/linearApproximation/exercises/LinearApproximations4.tex
@@ -26,18 +26,14 @@
 \]
 
 Using this linear approximation, approximate the value of $e$.
+
 \begin{hint}
-We have to express $e$ as $f(x)$, for some $x$.
-\end{hint}
-\begin{hint}
-It follows that $e=e^1=e^{2(\frac{1}{2})}=f\left(\frac{1}{2}\right)$.
-\end{hint}
-\begin{hint}
-It follows that $e=f\left(\frac{1}{2}\right)\approx L(\frac{1}{2})$.
+Note that $e=e^1=e^{2(\frac{1}{2})}=f\left(\frac{1}{2}\right)$.
 \end{hint}
+
 \[
 e \approx \answer{2}.
 \]
 
 \end{exercise}
-\end{document}
\ No newline at end of file
+\end{document}