daviddoji/CalculusWithJuliaNotes.jl
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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)
``` ```
@ -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)?