diff --git a/quarto/derivatives/mean_value_theorem.qmd b/quarto/derivatives/mean_value_theorem.qmd
index 5eab0f8..21d9d5a 100644
--- a/quarto/derivatives/mean_value_theorem.qmd
+++ b/quarto/derivatives/mean_value_theorem.qmd
@@ -24,10 +24,10 @@ nothing
 ---
 
 
-A function is *continuous* at $c$ if $f(c+h) - f(c) \rightarrow 0$ as $h$ goes to $0$. We can write that as $f(c+h) - f(x) = \epsilon_h$, with $\epsilon_h$ denoting a function going to $0$ as $h \rightarrow 0$. With this notion, differentiability could be written as $f(c+h) - f(c) - f'(c)h = \epsilon_h \cdot h$.  This is clearly a more demanding requirement that mere continuity at $c$.
+A function is *continuous* at $c$ if $f(c+h) - f(c) \rightarrow 0$ as $h$ goes to $0$. We can write that as $f(c+h) - f(c) = \epsilon_h$, with $\epsilon_h$ denoting a function going to $0$ as $h \rightarrow 0$. With this notion, differentiability could be written as $f(c+h) - f(c) - f'(c)h = \epsilon_h \cdot h$.  This is clearly a more demanding requirement than mere continuity at $c$.
 
 
-We defined a function to be *continuous* on an interval $I=(a,b)$ if it was continuous at each point $c$ in $I$. Similarly, we define a function to be *differentiable* on the interval $I$ it it is differentiable at each point $c$ in $I$.
+We defined a function to be *continuous* on an interval $I=(a,b)$ if it was continuous at each point $c$ in $I$. Similarly, we define a function to be *differentiable* on the interval $I$ if it is differentiable at each point $c$ in $I$.
 
 
 This section looks at properties of differentiable functions. As there is a more stringent definition, perhaps more properties are a consequence of the definition.
@@ -55,7 +55,7 @@ f(x) = abs(x)
 plot(f, -1,1)
 ```
 
-We can see formally that the secant line expression will not have a limit when $c=0$ (the left limit is $-1$, the right limit $1$). But more insight is gained by looking a the shape of the graph.  At the origin, the graph always is vee-shaped. There is no linear function that approximates this function well. The function is just not smooth enough, as it has a kink.
+We can see formally that the secant line expression will not have a limit when $c=0$ (the left limit is $-1$, the right limit $1$). But more insight is gained by looking at the shape of the graph.  At the origin, the graph always is vee-shaped. There is no linear function that approximates this function well. The function is just not smooth enough, as it has a kink.
 
 
 There are other functions that have kinks. These are often associated with powers. For example, at $x=0$ this function will not have a derivative:
@@ -108,7 +108,7 @@ A hiker can appreciate the difference. A relative maximum would be the crest of
 What does this have to do with derivatives?
 
 
-[Fermat](http://science.larouchepac.com/fermat/fermat-maxmin.pdf), perhaps with insight from Kepler, was interested in maxima of polynomial functions. As a warm up, he considered a line segment $AC$ and a point $E$ with the task of choosing $E$ so that $(E-A) \times (C-A)$ being a maximum. We might recognize this as finding the maximum of $f(x) = (x-A)\cdot(C-x)$ for some $A < C$. Geometrically, we know this to be at the midpoint, as the equation is a parabola, but Fermat was interested in an algebraic solution that led to more generality.
+[Fermat](http://science.larouchepac.com/fermat/fermat-maxmin.pdf), perhaps with insight from Kepler, was interested in maxima of polynomial functions. As a warm up, he considered a line segment $AC$ and a point $E$ with the task of choosing $E$ so that $(E-A) \times (C-E)$ being a maximum. We might recognize this as finding the maximum of $f(x) = (x-A)\cdot(C-x)$ for some $A < C$. Geometrically, we know this to be at the midpoint, as the equation is a parabola, but Fermat was interested in an algebraic solution that led to more generality.
 
 
 He takes $b=AC$ and  $a=AE$. Then the product is $a \cdot (b-a) = ab - a^2$. He then perturbs this writing $AE=a+e$, then this new product is $(a+e) \cdot (b - a - e)$. Equating the two, and canceling like terms gives $be = 2ae + e^2$. He cancels the $e$ and basically comments that this must be true for all $e$ even as $e$ goes to $0$, so $b = 2a$ and the value is at the midpoint.
@@ -548,13 +548,13 @@ numericq(c)
 ###### Question
 
 
-The extreme value theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $f(x) = \sin(x)$ on $I=[\pi, 3\pi/2]$ find a value $c$ satisfying the theorem for an absolute maximum.
+The extreme value theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $f(x) = \cos(x)$ on $I=[\pi, 3\pi/2]$ find a value $c$ satisfying the theorem for an absolute maximum.
 
 
 ```{julia}
 #| hold: true
 #| echo: false
-c = pi
+c = 1pi
 numericq(c)
 ```