Merge pull request #79 from fangliu-tju/main

Update first_second_derivatives.qmd

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
@@ -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 $t_0$: