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Merge pull request #97 from jverzani/v0.18

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adjust question
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john verzani 2023-05-23 10:55:25 -04:00 committed by GitHub
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quarto/derivatives/mean_value_theorem.qmd
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@ -548,13 +548,13 @@ numericq(c)
###### Question ###### Question
The extreme value theorem is a guarantee of a value, but does not provide a recipe to find it. For the function $f(x) = \cos(x)$ 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 $c$ in $I$ for which $f(x)$ has its maximum value.
```{julia} ```{julia}
#| hold: true #| hold: true
#| echo: false #| echo: false
c = 1pi c = 3pi/2
numericq(c) numericq(c)
``` ```
@ -628,7 +628,7 @@ Suppose $f''(x) > 0$ on $I$. Why is it impossible that $f'(x) = 0$ at more than
choices = [ choices = [
L"It isn't. The function $f(x) = x^2$ has two zeros and $f''(x) = 2 > 0$", 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", "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." L"By the mean value theorem, we must have $f'(b) - f'(a) > 0$ whenever $b > a$. This means $f'(x)$ is increasing and can't double back to have more than one zero."
] ]
answ = 3 answ = 3
radioq(choices, answ) radioq(choices, answ)
@ -675,7 +675,7 @@ radioq(choices, answ)
###### Question ###### Question
In an example, we used the fact that if $0 < c < x$, for some $c$ given by the mean value theorem and $f(x)$ goes to $0$ as $x$ goes to zero then $f(c)$ will also go to zero. Suppose we say that $c=g(x)$ for some function $c$. In an example, we used the fact that if $0 < c < x$, for some $c$ given by the mean value theorem and $f(x)$ goes to $0$ as $x$ goes to zero then $f(c)$ will also go to zero. As $c$ depends on $x$, suppose we write $c=g(x)$ for some function $g$.
Why is it known that $g(x)$ goes to $0$ as $x$ goes to zero (from the right)? Why is it known that $g(x)$ goes to $0$ as $x$ goes to zero (from the right)?