@@ -548,13 +548,13 @@ numericq(c)
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###### Question
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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.
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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.
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```{julia}
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#| hold: true
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#| echo: false
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c = 1pi
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c = 3pi/2
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numericq(c)
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```
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@@ -675,7 +675,7 @@ radioq(choices, answ)
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###### Question
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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 $c$.
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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 $g$.
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Why is it known that $g(x)$ goes to $0$ as $x$ goes to zero (from the right)?
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