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calculus:resources:calculus_flipped_resources:applications:closed_interval_method_tex

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\begin{document} \begin{frame} \large Let $f(x)$ be a differentiable function on a closed interval with $x=a$ being one of the endpoints of the interval. If $f'(x)>0$ for all $x$, then \vskip 15pt \begin{itemize} \item[\bf (a)] $f$ could have either an absolute maximum or minimum at $x=a$. \vskip 15pt \item[\bf (b)] $f$ cannot have an absolute maximum at $x=a$. \vskip 15pt \item[\bf (c)] $f$ must have an absolute minimum at $x=a$. \vskip 15pt \item[\bf (d)] $x=a$ must be a critical number for $f$. \end{itemize} \end{frame} \begin{frame} \large If $f$ is continuous on $[a,b]$, then \vskip 15pt \begin{itemize} \item[\bf (a)] there must be local extreme values, but there may or may not be an absolute maximum or minimum value for the function. \vskip 15pt \item[\bf (b)] there must be numbers $m$ and $M$ such that $m\leq f(x) \leq M$, for all $x$ in $[a,b]$. \vskip 15pt \item[\bf (c)] any absolute maximum or minimum would be at either the endpoints of the interval, or at places in the domain where $f'(x)=0$. \end{itemize} \end{frame} \begin{frame} \large Find the absolute extrema of: \vskip 15pt \begin{itemize} \item[\bf (a)] $f(x)=x^3-3x+1$ on the interval $[0,3]$. \vskip 15pt \item[\bf (b)] $g(x)=\dfrac{x^2-4}{x^2+4}$ on the interval $[-4,4]$. \vskip 15pt \item[\bf (c)] $h(t)=t\sqrt{4-t^2}$ on the interval $[-1,2]$. \vskip 15pt \item[\bf (d)] $i(x)=x+\cot\left(\frac{x}{2}\right)$ on the interval $\left[\frac{\pi}{4},\frac{7\pi}{4}\right]$. \end{itemize} \end{frame} \begin{frame} \large Find the highest and lowest points on the graph of $f(x) = x^{3}-3x+6$ on the following intervals: \vskip 15pt \begin{itemize} \item[\bf (i)] $[-2,2]$. \vskip 15pt \item[\bf (ii)] $[-2,3]$. \vskip 15pt \item[\bf (iii)] $(-2,3)$. \end{itemize} \end{frame} \begin{frame} \large Show that the maximum and minimum values of the function $$f(x) = x^{3}+ax^{2}+bx+c$$ on the interval $[p,q]$ occur at the endpoints if $a^{2} <3b$. \vskip 50pt If $a$ and $b$ are positive numbers, find the maximum value of $$f(x)=x^a(1-x)^b$$ on the interval $[0,1]$. \end{frame} \end{document}
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