use quarto, not Pluto to render pages

CwJ/derivatives/mean_value_theorem.jmd

@@ -14,7 +14,7 @@ using CalculusWithJulia.WeaveSupport
 using Printf
 using SymPy
 
-fig_size = (600, 400)
+fig_size = (800, 600)
 
 const frontmatter = (
         title = "The mean value theorem for differentiable functions.",
@@ -84,19 +84,14 @@ power rule: $f'(x) = 1/3 \cdot x^{-2/3}$, which has a vertical
 asymptote at $x=0$.
 
 
-```julia; echo=false
-note("""
-
-The `cbrt` function is used above, instead of `f(x) = x^(1/3)`, as the
-latter is not defined for negative `x`. Though it can be for the exact
-power `1/3`, it can't be for an exact power like `1/2`. This means the
-value of the argument is important in determining the type of the
-output - and not just the type of the argument. Having type-stable
-functions is part of the magic to making `Julia` run fast, so `x^c` is
-not defined for negative `x` and most floating point exponents.
-
-""")
-```
+!!! note
+	The `cbrt` function is used above, instead of `f(x) = x^(1/3)`, as the
+	latter is not defined for negative `x`. Though it can be for the exact
+	power `1/3`, it can't be for an exact power like `1/2`. This means the
+	value of the argument is important in determining the type of the
+	output - and not just the type of the argument. Having type-stable
+	functions is part of the magic to making `Julia` run fast, so `x^c` is
+	not defined for negative `x` and most floating point exponents.
 
 
 Lest you think that continuous functions always have derivatives
@@ -128,14 +123,9 @@ must also be specified, for a relative maximum there just needs to
 exist some interval, possibly really small, though it must be bigger
 than a point.
 
-```julia; echo=false
-note("""
-
-A hiker can appreciate the difference. A relative maximum would be the
-crest of any hill, but an absolute maximum would be the summit.
-
-""")
-```
+!!! note
+	A hiker can appreciate the difference. A relative maximum would be the
+	crest of any hill, but an absolute maximum would be the summit.
 
 What does this have to do with derivatives?
 
@@ -286,19 +276,14 @@ Here the maximum occurs at an endpoint. The critical point $c=0.67\dots$
 does not produce a maximum value. Rather $f(0.67\dots)$ is an absolute
 minimum.
 
-```julia; echo=false
-note(L"""
-
-**Absolute minimum** We haven't discussed the parallel problem of
+!!! note
+  **Absolute minimum** We haven't discussed the parallel problem of
   absolute minima over a closed interval. By considering the function
   $h(x) = - f(x)$, we see that the any thing true for an absolute
   maximum should hold in a related manner for an absolute minimum, in
   particular an absolute minimum on a closed interval will only occur
   at a critical point or an end point.
 
-""")
-```
-
 ## Rolle's theorem
 
 Let $f(x)$ be differentiable on $(a,b)$ and continuous on
@@ -616,8 +601,8 @@ choices = [
 "``h(x) = f(x) - g(x)``",
 "``h(x) = f'(x) - g'(x)``"
 ]
-ans = 3
-radioq(choices, ans)
+answ = 3
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -630,8 +615,8 @@ L"It isn't. The function $f(x) = x^2$ has two zeros and $f''(x) = 2 > 0$",
 "By the Rolle's theorem, there is at least one, and perhaps more",
 L"By the mean value theorem, we must have $f'(b) - f'(a) > 0$ when ever $b > a$. This means $f'(x)$ is increasing and can't double back to have more than one zero."
 ]
-ans = 3
-radioq(choices, ans)
+answ = 3
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -645,8 +630,8 @@ choices = [
 "``c = 1 / (1/a + 1/b)``",
 "``c = a + (\\sqrt{5} - 1)/2 \\cdot (b-a)``"
 ]
-ans = 2
-radioq(choices, ans)
+answ = 2
+radioq(choices, answ)
 ```
 
 ###### Question
@@ -660,8 +645,8 @@ choices = [
 "``c = 1 / (1/a + 1/b)``",
 "``c = a + (\\sqrt{5} - 1)/2 \\cdot (b-a)``"
 ]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 
@@ -678,8 +663,8 @@ Why is it known that $g(x)$ goes to $0$ as $x$ goes to zero (from the right)?
 choices = [L"The squeeze theorem applies, as $0 < g(x) < x$.",
 L"As $f(x)$ goes to zero by Rolle's theorem it must be that $g(x)$ goes to $0$.",
 L"This follows by the extreme value theorem, as there must be some $c$ in $[0,x]$."]
-ans = 1
-radioq(choices, ans)
+answ = 1
+radioq(choices, answ)
 ```
 
 Since $g(x)$ goes to zero, why is it true that if $f(x)$ goes to $L$ as $x$ goes to zero that $f(g(x))$ must also have a limit $L$?
@@ -688,6 +673,6 @@ Since $g(x)$ goes to zero, why is it true that if $f(x)$ goes to $L$ as $x$ goes
 choices = ["It isn't true. The limit must be 0",
 L"The squeeze theorem applies, as $0 < g(x) < x$",
 "This follows from the limit rules for composition of functions"]
-ans = 3
-radioq(choices, ans)
+answ = 3
+radioq(choices, answ)
 ```