Title: Math 2250 Chapter 4 (Applications of Derivatives)

Q: [latex]
The \emph{absolute maximum value} or \emph{global maximum value} of a function $f(x)$ on an interval $[a,b]$ is
[/latex]
A: [latex]
The value $f(x)$ with $x$ in $[a,b]$ so that if $c$ is any number in $[a,b]$,
\begin{equation*}
f(x) \geq c
\end{equation*}
Note: There may be more than one $x$ where $f(x)$ is the absolute maximum value of $f(x)$.
[/latex]

Q: [latex]
The \emph{absolute minimum value} or \emph{global minimum value} of $f(x)$ on an interval $[a,b]$ is
[/latex]
A: [latex]
the value $f(x)$ with $x$ in $[a,b]$ so that if $c$ is any number in $[a,b]$,
\begin{equation*}
f(x) \leq f(c)
\end{equation*}
Note: There may be more than one $x$ where $f(x)$ is the absolute minimum value of $f$ on $[a,b]$.
[/latex]

Q: [latex]
A function $f(x)$ has a \emph{local maximum value} $f(x)$ at some $x$ if
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A: [latex]
There is SOME open interval $(c,d)$ containing $x$ so that
\begin{equation*}
f(x) \geq f(y)
\end{equation*}
for all $y \in (c,d)$.
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[latex]
Note. There may be $y$ in $[a,b]$ (but outside the smaller interval $(c,d)$) so that $f(y) > f(x)$.
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Q: [latex]
A function $f(x)$ defined on $[a,b]$ has a \emph{local minimum value} $f(x)$ at $x$ in $[a,b]$ if
[/latex]
A: [latex]
there is SOME interval $(c,d)$ containing $x$ so that
\begin{equation*}
f(x) \geq f(y)
\end{equation*}
for all $y \in (c,d)$.
Note: There can be $y$ in $[a,b]$ (but outside $(c,d)$) where $f(y) < f(x)$.
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Q: [latex]
If $f(x)$ has a local maximum or minimum value at $c$ in $(a,b)$ then
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A: [latex]
the derivative of $f(x)$ vanishes at $c$:
\begin{equation*}
f'(c) = 0
\end{equation*}
Notes.
\begin{enumerate}
\item Just because $f'(c) = 0$ doesn't automatically mean that $f(c)$ is a local maximum or minimum value.
\item Remember that a \emph{local} maximum or minimum value is not always the \emph{absolute} maximum or minimum value. There are lots of local maxes and mins, but only one absolute max \emph{value} and absolute min \emph{value}.
\end{enumerate}
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Q: [latex]
A \emph{critical point} of a function $f(x)$ defined on $[a,b]$ is
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A: [latex]
an \emph{interior point} $x$ of the domain (that is, a point in the open interval $(a,b)$) where either
\begin{enumerate}
\item $f'(x) = 0$ or
\item $f'(x)$ doesn't exist
\end{enumerate}
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Q: [latex]
To find the absolute max and min of $f(x)$ on $[a,b]$, we
[/latex]
A: [latex]
\begin{enumerate}
\item Evaluate $f(x)$ at all critical points and on the endpoints $a$ and $b$.
\item Take the largest and smallest of these values.
\end{enumerate}
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Q: [latex]
Rolle's Theorem is
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A: [latex]
If the derivative of $f(x)$ exists at every point in $[a,b]$ and $f(a) = f(b)$ then there is some $c$ in $(a,b)$ so that $f'(c) = 0$.
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Q: [latex]
The Mean Value Theorem is
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A: [latex]
Suppose that the derivative of $f(x)$ exists at every point in $[a,b]$. Then there is at least one point $c \in (a,b)$ where the derivative $f'(c)$ is equal to the average rate of change of $f$ over $[a,b]$.
\begin{equation*}
\frac{f(b) - f(a)}{b - a} = f'(c).
\end{equation*}
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Q: [latex]
If the derivative $f'(x)$ is \emph{positive} ($f'(x) > 0$) on $[a,b]$
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A: [latex]
the function is \emph{increasing} on $[a,b]$ (and $f(b) > f(a)$).
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Q: [latex]
If the derivative $f'(x)$ is \emph{negative} ($f'(x) < 0$) on $[a,b]$
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A: [latex]
then $f(x)$ is \emph{decreasing} on $[a,b]$ (and $f(b) < f(a)$).
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Q: [latex]
If $f'(c) = 0$ and $f'(x)$ switches \emph{from positive to negative} at $c$ as $x$ increases
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A: [latex]
then $f(c)$ is a \emph{local maximum for $f(x)$}
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Q: [latex]
If $f'(c) = 0$ and $f'(x)$ switches \emph{from negative to positive} at $c$ as $x$ increases
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A: [latex]
then $c$ is a local \emph{minimum} for $f$.
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Q: [latex]
A function $f(x)$ is \emph{concave up} on $[a,b]$ if
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A: [latex]
the second derivative $f''(x) > 0$ on $[a,b]$.
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Q: [latex]
A function $f(x)$ is \emph{concave down} on $[a,b]$ if
[/latex]
A: [latex]
the second derivative $f''(x) < 0$ on $[a,b]$.
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Q: [latex]
(L'H\^ospital's Rule). If $f(a) = g(a) = 0$ and $f'(a)$ and $g'(a)$ exist and $g'(a)$ is not zero, then
\begin{equation*}
\lim_{x \rightarrow a} \frac{f(a)}{g(a)} =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}\lim_{x \rightarrow a} \frac{f(a)}{g(a)} = \lim_{x \rightarrow a} \frac{f'(a)}{g'(a)}
\end{equation*}
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Q: [latex]
Newton's method for solving $f(x) = 0$ is
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A: [latex]
\begin{enumerate}
\item Guess some value $x_0$ where you think $f(x_0)$ is pretty close to zero.
\item Use this approximation to get another approximation $x_1$ (and that approximation to get another approximation $x_2$, and so forth) using the formula
\begin{equation*}
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\end{equation*}
\end{enumerate}
Note: This can fail to converge to the right answer if the initial value $x_0$ is very badly chosen. If things don't look good by the 3rd or 4th step, give up and choose a better starting point.
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Q: [latex]
A function $F(x)$ is an antiderivative of $f(x)$ if
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A: [latex]
$F'(x) = f(x)$.
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Q: [latex]
\begin{equation*}
\int x^n \, dx =
\end{equation*}
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A: [latex]
\begin{equation*}
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \sin k x \, dx =
\end{equation*}
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A: [latex]
\begin{equation*}
\int \sin k x \, dx = -\frac{1}{k} \cos k x + C
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \cos k x \, dx =
\end{equation*}
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A: [latex]
\begin{equation*}
\int \cos k x \, dx = \frac{1}{k} \sin k x + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \sec^2 k x \, dx =
\end{equation*}
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A: [latex]
\begin{equation*}
\int \sec^2 k x \, dx = \frac{1}{k} \tan k x + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \csc^2 k x \, dx=
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \csc^2 k x \, dx = - \frac{1}{k} \cot k x + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \sec kx \tan kx \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \sec kx \tan kx \, dx = \frac{1}{k} \sec kx + C.
\end{equation*}
[/latex]

Q: [latex]
\begin{equation*}
\int \csc kx \cot kx \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \csc kx \cot kx \, dx = - \frac{1}{k} \csc kx + C.
\end{equation*}
[/latex]

Q: [latex]
\begin{equation*}
\int e^{kx} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \frac{1}{x} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \frac{1}{x} \, dx = \ln |x| + C.
\end{equation*}
The absolute value might look weird here, but trying an example should convince you that it's right. If
\begin{equation*}
|x| = \begin{cases}
x & \text{if } x > 0 \\
-x & \text{if } x < 0 \\
\end{cases}
\end{equation*}
then for $x < 0$, we see that
\begin{equation*}
\frac{d}{dx} \ln |x| = \frac{d}{dx} \ln -x = -\frac{1}{-x} = \frac{1}{x}
\end{equation*}
by the chain rule (and $\ln |x|$ is defined for $x < 0$ while $\ln x$ isn't).
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Q: [latex]
\begin{equation*}
\int \frac{1}{\sqrt{1 - k^2 x^2}} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \frac{1}{\sqrt{1 - k^2 x^2}} \, dx = \frac{1}{k} \arcsin k x + C.
\end{equation*}
[/latex]

Q: [latex]
\begin{equation*}
\int \frac{1}{1 + k^2 x^2} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \frac{1}{1 + k^2 x^2} \, dx = \frac{1}{k} \arctan kx + C.
\end{equation*}
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Q: [latex]
\begin{equation*}
\int \frac{1}{x \sqrt{k^2 x^2 - 1}} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int \frac{1}{x \sqrt{k^2 x^2 - 1}} \, dx = \operatorname{arcsec} kx + C.
\end{equation*}
as long as $kx > 1$.
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Q: [latex]
\begin{equation*}
\int a^{kx} \, dx =
\end{equation*}
[/latex]
A: [latex]
\begin{equation*}
\int a^{kx} \, dx = \frac{1}{k \ln a} a^{kx} + C.
\end{equation*}
[/latex]