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, $L(9.5)\ge f(9.5)$.