Merge branch 'main' of https://github.com/jverzani/CalculusWithJuliaNotes.jl

quarto/derivatives/first_second_derivatives.qmd

@@ -184,7 +184,7 @@ Consider the function $f(x) = e^{-\lvert x\rvert} \cos(\pi x)$ over $[-3,3]$:
 plotif(𝐟, 𝐟', -3, 3)
 ```
 
-We can see the first derivative test in action: at the peaks and valleys – the relative extrema – the color changes. This is because $f'$ is changing sign as as the function changes from increasing to decreasing or vice versa.
+We can see the first derivative test in action: at the peaks and valleys – the relative extrema – the color changes. This is because $f'$ is changing sign as the function changes from increasing to decreasing or vice versa.
 
 
 This function has a critical point at $0$, as can be seen. It corresponds to a point where the derivative does not exist. It is still identified through `find_zeros`, which picks up zeros and in case of discontinuous functions, like `f'`, zero crossings:
@@ -220,7 +220,7 @@ scatter!(𝒇cps, 0*𝒇cps)
 We see the six zeros as stored in `cps` and note that at each the function clearly crosses the $x$ axis.
 
 
-From this last graph of the derivative we can also characterize the graph of $f$: The left-most critical point coincides with a relative minimum of $f$, as the derivative changes sign from negative to positive. The critical points then alternate relative maximum, relative minimum, relative maximum, relative, minimum, and finally relative maximum.
+From this last graph of the derivative we can also characterize the graph of $f$: The left-most critical point coincides with a relative minimum of $f$, as the derivative changes sign from negative to positive. The critical points then alternate relative maximum, relative minimum, relative maximum, relative minimum, and finally relative maximum.
 
 
 ##### Example
@@ -448,7 +448,7 @@ We won't work with these definitions in this section, rather we will characteriz
 A proof of this makes use of the same trick used to establish the mean value theorem from Rolle's theorem. Assume $f'$ is increasing and let $g(x) = f(x) - (f(a) + M \cdot (x-a))$, where $M$ is the slope of the secant line between $a$ and $b$. By construction $g(a) = g(b) = 0$. If $f'(x)$ is increasing, then so is $g'(x) = f'(x) + M$. By its definition above, showing $f$ is concave up is the same as showing $g(x) \leq 0$. Suppose to the contrary that there is a value where $g(x) > 0$ in $[a,b]$. We show this can't be. Assuming $g'(x)$ always exists, after some work, Rolle's theorem will ensure there is a value where $g'(c) = 0$ and $(c,g(c))$ is a relative maximum, and as we know there is at least one positive value, it must be $g(c) > 0$. The first derivative test then ensures that $g'(x)$ will increase to the left of $c$ and decrease to the right of $c$, since $c$ is at a critical point and not an endpoint. But this can't happen as $g'(x)$ is assumed to be increasing on the interval.
 
 
-The relationship between increasing functions and their derivatives – if $f'(x) > 0 $ on $I$, then $f$ is increasing on $I$ – gives this second characterization of concavity when the second derivative exists:
+The relationship between increasing functions and their derivatives – if $f'(x) > 0$ on $I$, then $f$ is increasing on $I$ – gives this second characterization of concavity when the second derivative exists:
 
 
 >   * If $f''(x)$ exists and is positive on $I$, then $f(x)$ is concave up on $I$.
@@ -522,7 +522,7 @@ We can check the sign of the second derivative for each critical point:
 [jcps j''.(jcps)]
 ```
 
-That $j''(0.6) < 0$ implies that at $0.6$, $j(x)$ will have a relative maximum. As $''(1) > 0$, the second derivative test says at $x=1$ there will be a relative minimum. That $j''(0) = 0$ says that only that there **may** be a relative maximum or minimum at $x=0$, as the second derivative test does not speak to this situation. (This last check, requiring a function evaluation to be `0`,  is susceptible to floating point errors, so isn't very robust as a general tool.)
+That $j''(0.6) < 0$ implies that at $0.6$, $j(x)$ will have a relative maximum. As $j''(1) > 0$, the second derivative test says at $x=1$ there will be a relative minimum. That $j''(0) = 0$ says that only that there **may** be a relative maximum or minimum at $x=0$, as the second derivative test does not speak to this situation. (This last check, requiring a function evaluation to be `0`,  is susceptible to floating point errors, so isn't very robust as a general tool.)
 
 
 This should be consistent with this graph, where $-0.25$, and $1.25$ are chosen to capture the zero at $0$ and the two relative extrema:
@@ -979,7 +979,7 @@ answ=1
 radioq(choices, answ)
 ```
 
-##### Question
+###### Question
 
 
 You know $f''(x) = (x-1)^3$. What do you know about $f(x)$?
@@ -997,7 +997,7 @@ answ = 1
 radioq(choices, answ)
 ```
 
-##### Question
+###### Question
 
 
 While driving we accelerate to get through a light before it turns red. However, at time $t_0$ a car cuts in front of us and we are forced to break. If $s(t)$ represents position, what is $t_0$: