diff --git a/meanValueTheorem/digInMeanValueTheorem.tex b/meanValueTheorem/digInMeanValueTheorem.tex
index 9ab6f39..8d3a2f7 100644
--- a/meanValueTheorem/digInMeanValueTheorem.tex
+++ b/meanValueTheorem/digInMeanValueTheorem.tex
@@ -298,7 +298,7 @@
 		We calculate $f(0) = \answer[given]{1}$ and $f(3)=\answer[given]{e^3}$. That means the average rate of change of $f$
 		on the interval $[0, 3]$ is $\dfrac{f(3)-f(0)}{3-0}=\answer[given]{\dfrac{e^3-1}{3}}$.
 
-		The value of $c$ will be in the interval $(0, 3)$ will satisfy $f'(c) = \dfrac{e^3-1}{3}$.
+		The value of $c$ will be in the interval $(0, 3)$ and will satisfy $f'(c) = \dfrac{e^3-1}{3}$.
 		Since $f'(x) = e^x$, this means $c$ will be the solution to the equation:
 		$\displaystyle  e^c = \dfrac{e^3-1}{3}$. Taking the natural logarithm of both sides of this equation gives:
 		\[c = \ln\left( \dfrac{e^3-1}{3} \right). \]
@@ -312,7 +312,7 @@
 In conclusion, the Mean Value Theorem relates the function $f$ and its derivative, $f'$. Since the derivative has many interpretations,
  e.g. instantaneous rate of change, slope of the tangent line, velocity, it is no surprise that we can use the MVT in different contexts.
 
-Therefore, if the the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the MVT guarantees that
+Therefore, if the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the MVT guarantees that
 \begin{enumerate}
 	\item there is a point $c$ in $(a,b)$ where the instantaneous rate of change of $f$ is equal to the average 
 		rate of change of $f$ over the interval $[a,b]$;