Job 17 monotonicity

1. Solution: Function $y = (x-1) \cdot \sqrt[3]{x^2}$ within the defined domain $ (-\infty,+\infty) $, and
\[
F ' (x) = \frac53 x^{2/3}-\frac23 X^{-1/3}.
\]
Make $f ' (x) =0$ $x =\frac25$, and notice that when $x =0$, the function is not directed. The list is examined as follows:
\begin{center}
\begin{tabular}{|c|c |c | c | c|c |}
\hline
$x $ & $ (-\infty,0) $ & 0& $ (0,\frac25) $ & \mbox{$\frac25$} & $ (\frac25,+\infty) $ \\[2ex]
$f ' (x) $ & + & \mbox{not present} & $-$ & 0 & + \\[2ex]
$f (x) $ & $\nearrow$ & \mbox{Max} & $\searrow$ & \mbox{min} & $\nearrow$ \
\hline
\end{tabular}
\end{center}
Therefore, the monotonically increasing interval of the function is $ (-\infty,0], [\frac25,+\infty) $, and the monotone minus interval is $[0,\frac25]$. The maximum value is $f (0) =-1$, and the minimum value is $f (\frac25) =-\frac15 (\frac{2}{5}) ^{2/3}$.



2. First,
\[
f (x) =
\begin{cases}
X E^{1-x}, & X\geq 1,\\
X e^{x-1}, & 0<x< 1, \ \
-X E^{x-1}, & X\leq 0.
\end{cases}
\]
By definition, it is easy to verify that the $f (x) $ is not =1$ at the staging point $x =0$ and at $x, so
\[
F ' (x) =
\begin{cases}
E^{1-x}-xe^{1-x}, & X> 1,\\
\mbox{not exist}, & x=1,\\
E^{x-1}+x e^{x-1}, & 0<x< 1, \ \
\mbox{not exist}, & x=0,\\
-e^{x-1}-xe^{x-1}, & x< 0.
\end{cases}
\]
Make $f ' (x) =0$ $x =-1$. The list is discussed below:
\begin{center}
\begin{tabular}{|c|c|c |c|c | c | c|c |}
\hline
$x $ & $ (-\infty,-1) $ & -1& $ ( -1,0) $ & 0 & $ (0,1) $ & 1& $ (1,+\infty) $ \\[2ex]
$f ' (x) $ & + & 0 & $-$ & \mbox{Not present} & + & \mbox{not present} & $-$ \\[2ex]
$f (x) $ & $\nearrow$ & \mbox{Max} & $\searrow$ & \mbox{min} & $\nearrow$ & \mbox{Max}& $\searro w$\\
\hline
\end{tabular}
\end{center}
Therefore, the function $f (x) $ has a maximum value $f ( -1) =e^{-2}, F (1) =1$ and minimum $f (0) =0$.



3. Solution: First,
\[
Y ' = \frac{1}{x^2+1}.
\]
So when $0\leq x \leq 1$ $y ' >0$, so the minimum $y (0) =-\frac{\pi}{4}$, the maximum $y (1) =0$.



4. Proof: First make
\[
\phi (x) =\sin x-x,
\]
When $0<x<\pi/2$,
\[
\phi ' (x) =\cos x-1<0,
\]
So
\[
\sin x-x=\phi (x) <\phi (0) = 0,
\]
That
\[
\sin x < x.
\]
Again make
\[
\PSI (x) = \sin x-\frac{2}{\pi} x,
\]
The
\[
\psi ' (x) =\cos x-\frac 2\pi=0
\]
Solution $x = \arccos \frac 2\pi$. and
\[
\psi "(\arccos \frac{2}{\pi})
=
-\sin (\frac{2}{\pi}) <0.
\]
So $x = \arccos \frac 2\pi$, $x =0$ and the possible maximum points for $x =\frac{\pi}{2}$, and $x = \arccos \frac 2\pi$ is the largest point, so
\[
\sin x-\frac2 \pi x > \psi (0) = 0,
\]
That is, when $0<x<\frac \pi 2$,
\[
\frac 2\pi x <\sin x <x.
\]


5. Proof: According to the question, when the $x \neq 0$,
\[
F ' (x) =
% \begin{cases}
\frac{XF ' (x)-F (x)}{x^2},
% & X\neq 0,\\
% \frac{f "(0)}{2}, & X=0.
%\end{cases}
\]
Make
\[
\phi (x) =xf ' (x)-F (x),
\]
and
\[
\phi ' (x) =xf "(x).
\]
When $x >0$, $\phi ' (x) >0$, so $\phi (x) $ on $[0,+\infty) $ monotonically increases, so
\[
\phi (x) >\phi (0) =0.
\]
That is, when the $x >0$, $F ' (x) >0$, so $F (x) $ on $[0,+\infty) $ on the monotonically increasing.

On the other hand, when the $x <0$, $\phi ' (x) <0$, so $\phi (x) $ on $ (-\infty,0]$ monotonically minus, so
\[
\phi (x) >\phi (0) =0.
\]
That is, when $x <0$, $F ' (x) >0$, so $F (x) $ is monotonically increasing on $ (-\infty,0]$).
In summary, $F (x) $ is a monotonically increasing function.



6. Solution: First notice that the defined domain of the $x \ln x$ is $ (0,+\infty) $, and
\[
(X\ln x) ' =\ln x+1,
\qquad
(X\ln x) ' =\frac1x>0.
\]
So when the $x =1/e$, $x \ln x$ take the minimum $-1/e$, and when $0<x<1/e$, $x \ln x$ monotonically reduced, when $1/e<x$, $x \ln x$ monotonically increase. When the $0<x<1$, $x \ln x<0$, when $x >1$, $x \ln x>0$.

According to the topic, that is to discuss $x \ln x=-a$ have real roots.

(1) If the $a >1/e$, then because $x \ln x>-1/e$, so there is no real root.

(2) If $a =1/e$, there is a unique real root $x =1/e$.

(3) If $0<a<1/e$, because $\lim_{x\to 0}x\ln x=0$, it can be defined
\[
f (x) =
\begin{cases}
X\ln x, & x>0,\\
0, & X=0.
\end{cases}
\]
function $f (x) $ on $[0,+\infty) $ on continuous. In the case of $0\leq x \leq 1/e$, $-1/e \leq f (x) \leq 0$, according to the mean value theorem of the closed interval continuous function, there must be $x _1\in (0,1/e) $ to make $f (x_1) =x_1\ln x_1=-a$. Similarly, the existence of $x _2 \in (1/e,1) $ makes $f (x_2) =x_2\ln x_2=-a$, so there are two different real roots at this time.

(4) If $a =0$, $x _1=1$ is the only root.

(5) If the $a <0$, because $x >1$ when the monotonous increase, so there is only real roots.








7. Solution: Set Variable $x =h-r$, and $0<x<r$. The high $h =r+x$ of the positive cone, the base radius of the positive cone $r =\sqrt{r^2-x^2}$, the area of the positive cone is
\[
\frac13 \pi r^2 h=
\frac13 \pi (r^2-x^2) (r+x).
\]
and
\[
(r^2-x^2) (r+x) ' = (3x-r) (x+r) = 0,
\]
For $x = \frac R 3$ and $x =-r$, the maximum area is
\[
\frac13 \pi (r^2-(R/3) ^2) (R+R/3) = \frac{32}{81} \pi R.
\]

Note: Set the variable $x $ is the most convenient, if you consider the RADIUS $r $ as a variable, there is no way to solve the most value point, here the advantage of this intermediate variable is that the use of the positive cone formula is just the square root removed.

Job 17 monotonicity