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Fix writing errors #33

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Fix writing errors #33

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4 changes: 2 additions & 2 deletions meanValueTheorem/digInMeanValueTheorem.tex
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Expand Up @@ -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:
\[c = \ln\left( \dfrac{e^3-1}{3} \right). \]
Expand All @@ -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 $[a,b]$;
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