Job 14 differential mean value theorem

2. Proof: Set by the title, for any $x \in (A, A, b) $
\[f (x) G ' (x)-F ' (x) g (x) = 0, \]
So
$$ \left (\frac{f (x)}{g (x)} \right) ' =\frac{F ' (x) g (x)-F (x) G ' (x)}{g^2 (x)}=0, $$
Based on the differential mean value theorem inference
$$ \frac{f (x)}{g (x)}=k, $$
where $k $ is constant.





3. Proof: (1) Order
$$\varphi (x) =f (x)-x,$$
Easy to verify $\varphi (x) $ in $[\frac12, 1]$ on a continuous, in $ (\frac12,1) $ within the guide. Because
$$ \varphi (\FRAC12) \varphi (1) = (f (\FRAC12)-\FRAC12) (f (1)-1) =-\frac12<0, $$
According to the 0 point theorem, there is $c \in (\frac12,1) $ which makes $\varphi (c) =0 $, i.e.
$ $f (c) =c.$$


(2) for any real number $\lambda$, make
$$\PSI (x) = E^{-\lambda x} [F (x)-X]. $$
Easy to verify $\varphi (x) $ in $[0, c]$ on a continuous, in $ (0,c) $ within the guide. Because
$$ \psi (0) =0 $$
And
$$\PSI (c) = E^{-0\lambda c}[f (c)-c]=0, $$
According to the Lowe theorem, the existence of $\xi\in (0,C) $ makes
$$\psi ' (\xi) =-\lambda E^{-\lambda \xi} [F (\xi)-\xi]+e^{-\lambda \xi}[f ' (\xi) -1]=0.$$
Because $e ^{\lambda \xi}\neq 0$, both sides are multiplied by $e ^{\lambda \xi}$ to arrange
$ $f ' (\xi)-\lambda[f (\XI)-\xi]=1.$$



4. Certification: Launched according to the title
$$\frac{1}{\sqrt{x+1}+ \sqrt{x}}=\frac{1}{2\sqrt{x+\theta (x)}}$$
That
$$ 2\sqrt{x+\theta (x)}=\sqrt{x+1}+\sqrt{x}, $$
Therefore, the calculations are
$$ \theta (x) =\frac14 +\frac12 (\sqrt{x^2+x}-x). $$
Because
$$ 0\leq \sqrt{x^2+x}-x = \frac{x}{\sqrt{x^2+x}+x}\leq \frac{x}{x+x}=\frac12, $$
So
$$ \frac14 \leq \theta (x) \leq \frac12. $$
And because
$$\lim_{x\to 0} (\sqrt{x^2+x}-x) =0 \quad \mbox{and} \quad \lim_{x\to +\infty} (\sqrt{x^2+x}-x) =\lim_{x\to +\infty}\frac {x} {\sqrt{x^2+x}+x}=\lim_{x\to +\infty}\frac{1}{\sqrt{1+1/x}+1}=\frac12,$$
So
$$\lim_{x\to 0}\theta (x) =\frac14, \lim_{x\to +\infty}\theta (x) =\frac12.$$

Description: In fact, the bounds of $\theta (x) $ can be seen from a different perspective. When $x >0$
$$ (\sqrt{x^2+x}-x) ' = \frac{2x+1}{2\sqrt{x^2+x}}-1>0, $$
So $\theta (x) $ is the increment function, at this time in $x =0$, $\theta (x) $ Gets the minimum $\theta (0) =\frac14$, of course $\theta (x) \leq \lim_{x\to +\infty}\theta (x) =\f rac12$.




5.


(1) Proof: Because when $x \geq 1$
$$\left (\arccos \frac{2x}{1+x^2}\right) ' =-\frac{\frac{2 (1+x^2) -4x^2} {(1+x^2) ^2}} {\sqrt{1-\frac{(2x) ^2}{(1+x^2 ) ^2}}}=\frac{2}{1+x^2}, $$
That is, when the $x \geq 1$
$$\left (\arctan x-\frac12 \arccos \frac{2x}{1+x^2}\right) ' =0.$$
And because
$ \arctan 1-\frac12 \arccos \frac{2}{1+1}=\frac{\pi}{4}, $
So when $x \geq 1$,
$$ \arctan x-\frac12 \arccos \frac{2x}{1+x^2}=\frac{\pi}{4}. $$



(2) Proof: Books and courseware examples (ideas: proving monotonicity, and then proving the inequalities on either side, or using LaGrand-value theorem)




(3) Proof: According to the question set, namely the certificate
$$ n b^{n-1}< \frac{a^n-b^n}{a-b}<n a^{n-1}. $$
Make
$$\varphi (x) =x^n,$$
According to the LaGrand-value theorem, there is $b <\xi<a $ makes
$$\frac{a^n-b^n}{a-b}=\varphi ' (\xi) =n \xi^{n-1},$$ again because
$ $n b^{n-1}< n\xi^{n-1}<n a^{n-1}, $$
Certificate.



6. There is a proof of real roots using the 0-point theorem, only one real roots proof using the disprove method and the mean value theorem, see the courseware examples

Job 14 differential mean value theorem