calculus:resources:calculus_flipped_resources:applications:closed_interval_method_tex [Department of Mathematics and Statistics, Binghamton University]

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    \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}
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\begin{frame}
    \large Find the absolute extrema of:
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        \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}
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\begin{frame}
    \large Find the highest and lowest points on the graph of 
    \( f(x) = x^{3}-3x+6 \)  on the following intervals:
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        \begin{itemize}
            \item[\bf (i)]  \( [-2,2] \).
            \vskip 15pt

            \item[\bf (ii)]  \( [-2,3] \).
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            \item[\bf (iii)] \( (-2,3)\).
        \end{itemize}
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\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 $[0,1]$.
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