Exercise 18 convexity

Don't want to work today ... But I'm going to finish today's mission.

2. (1) Solution: function $y =e^{\arctan x}$ on the definition field $ (-\infty,+\infty) $ on a continuous, and
\[
Y ' = E^{\arctan x} \frac{1}{1+x^2}, \qquad
Y ' = E^{\arctan x} \frac{1}{(1+x^2) ^2}-e^{\arctan x}
\FRAC{2X}{1+X^2}.
\]
On $ (-\infty,+\infty) $ $y "$ exists, but does not keep the same symbol. Make $y ' =0$, $x =\frac12$. The list is examined as follows:
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$x $ & $ (-\INFTY,1/2) $ & $1/2$ & $ (+\infty) $ \ \
$f ' (x) $ & $-$ & $0$
& $+$ \ \
$f (x) $ & \mbox{on top}
& & \mbox{-convex}
\\
\hline
\end{tabular}
\end{center}
So, the function in $ (-\infty,1/2]$ is convex, in $[1/2,+\infty) $ is convex, the inflection point is $ (1/2,e^{\arctan \FRAC12}) $.



(2) Solution: function $y $ on the definition field $ (-\infty,+\infty) $ on a continuous, and
\[
Y ' =
\begin{cases}
1/x-1, & X\geq 1,
\\
2x-2, & X<1,
\end{cases}
\qquad
Y ' =
\begin{cases}
-1/x^2, & X>1,
\\
\mbox{does not exist}, & X=1,
\\
2, & X<1.
\end{cases}
\]
The list is examined as follows:
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$x $ & $ (-\infty,1) $ & $1$ & $ (1, +\infty) $ \ \
$f ' (x) $ & $+$ & \mbox{Not present}
& $-$ \ \
$f (x) $ & \mbox{}
& & \mbox{on Convex}
\\
\hline
\end{tabular}
\end{center}
So, the function in $ (-\infty,1]$ is convex, in $[1,+\infty) $ is convex, and the inflection point is $ (1,-1) $.



3.

(1) Proof: the Order function
\[
\phi (x) = X^n, \quad n>1,
\]
Then the function $\phi (x) $ on $[0,+\infty) $ on a continuous and at $ (0,+\infty) $ at least two orders can be directed. When $x >0$
\[
\phi ' (x) =n (n-1) x^{n-2}>0,
\]
launched when $x \in (0,+\infty) $ function $\phi (x) $ convex, that is, by definition for any $x >0,y>0$ have
\[
\frac{\phi (x) +\phi (y)} {2}>
\phi (\frac{x+y}{2}),
\]
That
\[
\frac{x^n +y^n}{2}
> \left (
\FRAC{X+Y}{2}
\right) ^n.
\]



(2) Proof: the Order function
\[
\phi (x) = X\ln x,
\]
The function $\phi (x) $ is contiguous on $ (0,+\infty) $. When $x >0$
\[
\phi ' (x) =\frac1x>0,
\]
launched when $x \in (0,+\infty) $ function $\phi (x) $ convex, that is, by definition for any $x _1>0,x_2>0$ have
\[
\frac{\phi (x_1) +\phi (x_2)} {2}>
\phi (\frac{x_1+x_2}{2}),
\]
In particular, taking $x _1=x,x_2=1$, the
\[
\frac{x\ln x +\ln 1}{2}
>
\FRAC{X+1}{2}
\ln\left (
\FRAC{X+1}{2}
\right)
\]
That
\[
{X\ln X}
>
({x+1})
\ln\left (
\FRAC{X+1}{2}
\right).
\]



4.

(1) Function $y =\frac{(x+1) ^3}{(x-1) ^2}$ in the definition domain $ (-\infty,1) \cup (1,+\infty) $ on continuous and
\[
Y ' = \frac{(x-5) (x+1) ^2}{(x-1) ^3},
\quad
Y ' = \frac{24 (x+1)} {(x-1) ^4},
\]
Make $y ' =0$ $x =-1,x=5$, so that $y ' =0$ $x =-1$. List discussion:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x $ & $ (-\infty,-1) $ & $-1$ & $ ( -1,1) $ & $1$ & $ (1,5) $ & & $ (5, +\infty) $ \ \
$f ' (x) $ & $+$ & $0$
& $+$ & \mbox{Not present} & $-$ & 0 &
$+$
\\
$f ' (x) $ & $-$ & $0$ & $+$
& \mbox{Not present} & $+$ & $+$ & $+$ \
$f (x) $ & $\nearrow$
& \mbox{inflection point, non-Extreme points} & $\nearrow$
& \mbox{Breaks} & $\searrow$ & \mbox{min}
& $\nearrow$
\\
\hline
\end{tabular}
\end{center}
In summary, the function has a minimum point of $5$, while $ ( -1,0) $ for its inflection points. On the other hand, the function has no horizontal asymptotic line, there are vertical asymptote $x =1$. and
\[
\lim_{x\to \infty}
\frac{(x+1) ^3}{x (x-1) ^2}=1,
\quad \lim_{x\to \infty}
[\frac{(x+1) ^3}{(x-1) ^2}-x]=5,
\]
So the function has oblique asymptote
\[
Y=x+5.
\]






(2) Function $y =x-2\arctan x$ on the definition field $ (-\infty,+\infty) $ on continuous and
\[
Y ' = 1-\frac{2}{1+x^2},
\quad
Y ' = \frac{4x}{(1+x^2) ^2},
\]
Make $y ' =0$ $x =\pm1$, so that $y ' =0$ $x =0$. List discussion:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x $ & $ (-\infty,-1) $ & $-1$ & $ ( -1,0) $ & $0$ & $ (0,1) $ & & $ (1, +\infty) $ \ \
$f ' (x) $ & $+$ & $0$
& $-$ & $-$ & $-$ & 0 &
$+$
\\
$f ' (x) $ & $-$ & $-$ & $-$
& $0$ & $+$ & $+$ & $+$ \ \
$f (x) $ & $\nearrow$
& \mbox{Max} & $\searrow$
& \mbox{Inflection} & $\searrow$ & \mbox{min}
& $\nearrow$
\\
\hline
\end{tabular}
\end{center}
In summary, the function has a maximum point $-1$, and a minimum point $1$, while $ (0,0) $ for its inflection points. On the other hand, the function has no vertical asymptotic line and no horizontal asymptote. and
\[
\lim_{x\to +\infty}
\frac{x-2\arctan X}{x}=1,
\quad \lim_{x\to +\infty}
({X-2\arctan x}-x) =-\pi,
\]
And
\[
\lim_{x\to-\infty}
\frac{x-2\arctan X}{x}=1,
\quad \lim_{x\to-\infty}
({X-2\arctan x}-x) =\pi,
\]
So the function has oblique asymptote
\[
Y=x-\pi, Y=x+\pi.
\]







5. Proof: Because $f "(x_0) \neq 0$, it may be advisable to set $f" (X_0) >0$. So
\[
F ' ' (X_0)
=\lim_{x\to X_0}
\frac{f ' (x)-F ' (X_0)}{x-x_0}
=\lim_{x\to X_0}
\frac{f "(x)}{x-x_0}>0
\]
According to the limit of the local guarantee number, there is $\delta>0$, so when $0<|x-x_0|<\delta$
\[
\frac{f ' (x)}{x-x_0}>0,
\]
When $x _0<x<x_0+\delta$, $f "(x) >0$, and $x _0-\delta<x<x-0$, $f" (x) <0$, i.e. $ (x_0,f (X_0)) $ for inflection point.




Description: $x _0$ is not an extreme point at this time, why?

Exercise 18 convexity