From baffc6854bea5b8487a638ae26ba681a0c1aaa20 Mon Sep 17 00:00:00 2001
From: pfatheddin <156558883+pfatheddin@users.noreply.github.com>
Date: Sun, 24 Mar 2024 19:41:04 -0400
Subject: [PATCH] Update linearApproximation1.tex

https://ximera.osu.edu/mooculus/linearApproximation/exercises/exerciseList/linearApproximation/exercises/linearApproximation1

It is important for them to give the exact interval for underestimate or overestimate when they check the second derivative.
---
 linearApproximation/exercises/linearApproximation1.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/linearApproximation/exercises/linearApproximation1.tex b/linearApproximation/exercises/linearApproximation1.tex
index ec7be36fb..9622a0815 100644
--- a/linearApproximation/exercises/linearApproximation1.tex
+++ b/linearApproximation/exercises/linearApproximation1.tex
@@ -89,7 +89,7 @@
 \begin {hint}
 Recall: $f''(x)=-\frac{1}{4\cdot x^{\answer{\frac{3}{2}}}}$. 
 
-Since $f''(x)<0$ on $(0,\infty)$,  the graph of $f$ is concave down. Therefore, the tangent line lies above the graph of $f$, except at $x=9$, where $f(9)=L(9)$.
+Since $f''(x)<0$ on $\left(9,\answer{9.5}\right)$,  the graph of $f$ is concave down. Therefore, the tangent line lies above the graph of $f$, except at $x=9$, where $f(9)=L(9)$.
 
 
 This means that $L(x)\ge f(x)$. So, $L(9.5)\ge f(9.5)$.