339. Derivatives. Rolle's function.
by Virgil Nicula, Apr 5, 2012, 5:44 PM
I. Derivatives.
Prove that
.
.
.
For
prove that
.
.
For example,
.
Let
be real numbers, with
. Prove that
(see here).
Let
be a function
and a line
for which
.
For any
the tangent at
in the point
meets again the graph
in the point
. Prove that the point 
are collinear. In particular case, the extremum points (if exist) and the inflexion point of
are colinearly.
Proof. Prove easily that the points
are collinear
. The points
are collinear
But the points
are collinear, i.e.
.
Thus,
the points
are collinear.
Ascertain the range of the function
, where
.
Prove that for any
exists the inequality
.
Proof. Consider the function
, where
. Observe that
. Therefore,
, i.e.
. From circular permutations obtain still two relations. From the product of these three inequalities results the proposed inequality.
II. Exercises with Rolle's functions.
Consider a continue function
so that
. Prove that exists
such that
.
Proof. Let
. Observe that
is continue and ![$h(a)h(b)=f(a)f(b)\left[f^2(a)-1\right]\left[f^2(b)-1\right]<0$](//latex.artofproblemsolving.com/5/9/f/59fa9ce8557cb1a0079aa57ccdeae08047eb986c.png)
because
and
. Therefore, there is
so that
, i.e.
.
Consider a Rolle's function
for which
. Prove that
such that
.[/quote]
Proof. Construct the function
,
and
. Observe that
. Thus
the function
is a Rolle's function and
. Results that there is
such that
, i.e.
.
Let
be a Rolle's function , where
.
Prove that there are
,
such that
.
Proof. Construct the function
, where
. Observe that
is Rolle's function and
. From Rolle's theorem obtain that
exists
so that
, i.e.
. Also exists
so that
a.s.o.
Prove that for any
,
the equation
has only one zero
. Prove that
.
Proof. Denote
, where
. Prove easily that
. Thus
the function
is decreasing on
and
. Therefore exists and is unique
so that
.
Denote
and
, where
. Observe that
,
,
,
where
and
for any
. Let
be a neighbourhood of
. Observe that
. Thus exists
,
so that for any
we have
. For any neighbourhood
of
exists
so that for
we have
, i.e.
.
Consider a Rolle's function
for which
. Prove that for any
exists
so that
.
Given is the function
so that
,
.
Prove that exists
so that ![$(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$](//latex.artofproblemsolving.com/d/2/9/d292f5cecdee2a67e4753a7473148bdcaf2a9b9c.png)
Given is the function
so that
for which exists
so that
.
Prove that exists
such that
.
Given are
,
so that
,
and
. Prove that the function
,
where
is strict increasing and solve the equation
.
Solve in
the system of equations
.
Consider the function
, where
and
.
Prove that exists
so that
for any
and the equation
has at least one zero for any
.
Consider the function
, where
and
. Prove that the point
which given by the
Lagrange's theorem is uniquely and
. Deduct the inequalities 
Proof. The point
is given by relation
.Must prove
, i.e.
.
Denote
. Thus the relation
becomes
, where
. Therefore,

Can obtain the inequalities
,
from
thus : for the first put
,
, where
; for the second put
,
.
Prove that
for any
.
Proof. Since
obtain
for any

for any
. Apply the Lagrange's theorem to the function
on the segments
and
.
Therefore, exist
so that
and
. In conclusion,
for any
.
Let
be a polynomial with at least two roots
. Prove that for any real number
there is
such that
.
Proof. If
, then can choose
. Suppose
. In this case consider
, where
.
Observe that
is a Rolle function , i.e. a continue function in
,
and a differentiable function on
. Since
and
, from the Rolle's theorem obtain that there is
so that
, i.e.
.
Prove that the sum
is convergent (without inegral).
Proof. Consider
, where
and
. Observe that
. From the
Lagrange's theorem results that exists
so that
. Since the function
is increasing obtain
, i.e. 
. Our sequence is increasing and bounded
obtain
.
PP15. For a given
consider the function
, where
where
. Is it an injective function ? Find
.
Proof. Since
and the function
is continue we can apply the Darboux's property: for any
exist
for which
, i.e. our function isn't injectively. Define
, i.e.
. So


.
In conclusion, the function
is
on
and is
on
.
PP16. Find
, where 
Proof. Let
, where
. Observe that
. Define the relation
, i.e.
.Observe that
. Appear three cases :
, i.e.
.
, i.e.
and
for any

. Observe that 
we have
, i.e.
.
, i.e.
.
In conclusion,
, where
.
Therefore, the evolution of the function
is
and
.
See here and here






![$\left[\ln \left(x^2+1\right)\right]^{(n)}=2(-1)^n(n-1)!\cdot\frac {\cos\left(n\cdot \arctan\frac 1x\right)}{\sqrt {\left(x^2+1\right)^n}}$](http://latex.artofproblemsolving.com/d/b/c/dbc68e7b53668c889aa5f2071c9624bc5fd49bf9.png)


![$\left(\frac {1}{x^2+ax+b}\right)^{(n)}=\frac {2(-1)^nn!}{\sqrt {4b-a^2}}\cdot\frac {\sin\left[(n+1)\arctan\frac {\sqrt {4b-a^2}}{2x+a}\right]}{\sqrt {\left(x^2+ax+b\right)^{n+1}}}$](http://latex.artofproblemsolving.com/2/8/1/2812d6218813d8d96a160170a2162f7e2f423d44.png)


For example,


![$\left\{\begin{array}{c}
\{a,b,c\}\subset \left[0,\frac 43\right]\\\\
abc\le \frac{4}{27}\end{array}\right\|$](http://latex.artofproblemsolving.com/2/5/c/25c879394efeaa58bf13f8f421a080d98081a4ce.png)









For any






are collinear. In particular case, the extremum points (if exist) and the inflexion point of

Proof. Prove easily that the points







Thus,












![$\left[x^2(y+z)\right]^{y-x}\cdot\left[y^2(z+x)\right]^{z-y}\cdot\left[z^2(x+y)\right]^{x-z}\le 1$](http://latex.artofproblemsolving.com/4/f/1/4f16da2bc35b6be01ba36e0ba545759d478e338c.png)
Proof. Consider the function







II. Exercises with Rolle's functions.

![$f : [a,b]\rightarrow (-1,1)$](http://latex.artofproblemsolving.com/4/0/5/405f62dd0f0d69701030ec4b882e473b03f10833.png)



Proof. Let
![$h(x)=f(x)[1+f(a)f(b)]-[f(a)+f(b)]\ ,\ a\le x\le b$](http://latex.artofproblemsolving.com/8/d/b/8db8ace5e5d4aecb9f6de5a954e53b050c3e89e0.png)

![$h(a)h(b)=f(a)f(b)\left[f^2(a)-1\right]\left[f^2(b)-1\right]<0$](http://latex.artofproblemsolving.com/5/9/f/59fa9ce8557cb1a0079aa57ccdeae08047eb986c.png)
because






![$f : [0,1]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/8/8/c/88cbfe98fe75cfc9980f76712f68bb744eb41690.png)



Proof. Construct the function
![$G:[0,1]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/c/8/b/c8b92ef427a86940776c5925d850e3140d06e7ac.png)

![$G(x)=f(x)\cdot e^{-\frac{|m|}{x}},\ x\in (0,1]$](http://latex.artofproblemsolving.com/9/5/8/95877744de1c7fe785074898ec8324ba98dc64ce.png)

the function






![$f:[a,b]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/0/b/b/0bb1e0310b7cbff4a1d3f47a70e6925fca11a9ca.png)

Prove that there are



Proof. Construct the function
![$F:[a,b]\rightarrow\mathbb R$](http://latex.artofproblemsolving.com/1/7/b/17b050eace966b56c1a30d1cc2366ea51fc1ca85.png)
![$F(x)=\frac {(x-a)[f(a)-f(x)]}{x}$](http://latex.artofproblemsolving.com/e/0/d/e0de7f5aba9380af11ff4b777c2ebd83e814210b.png)


exists


![$c(a-c)f'(c)=a[f(c)-f(b)]$](http://latex.artofproblemsolving.com/4/e/d/4edd23004ccad9db1aebc9c90bc1148a497002d3.png)





![$\sqrt [n]{x+1}-\sqrt [n] x=\frac{1}{n}$](http://latex.artofproblemsolving.com/1/8/4/184c324d2fab61536d0ef20e53713a7d9e32050c.png)


Proof. Denote
![$f_n:[0,1]\rightarrow\mathbb R$](http://latex.artofproblemsolving.com/e/f/3/ef3c7803dd20e59ebd5f5ebbc4989a1136713c1a.png)
![$f(x)=\sqrt [n]{x+1}-\sqrt [n] x-\frac{1}{n}$](http://latex.artofproblemsolving.com/2/5/4/254cba68c87f1c9f9bf95b648f794ea42b285bbd.png)
![$f'(x)\ .a.s.\ \sqrt [n]{x^{n-1}}-\sqrt [n]{(x+1)^{n-1}}\ .a.s.\ x-(x+1)<0$](http://latex.artofproblemsolving.com/4/6/5/46533686bd1297e677eef8334a96cf5e00905fc8.png)
the function

![$(0,1]$](http://latex.artofproblemsolving.com/1/0/1/101678300d1aa11009468da8419aaf2c57b795bd.png)
![$f_n(0)=1-\frac 1n>0>\sqrt [n]2-\left(1+\frac 1n\right)=f_n(1)$](http://latex.artofproblemsolving.com/e/b/5/eb5179e06152a846eaa1043369511a74aab747ee.png)


Denote
![$g_n(x)=n\cdot\left(\sqrt [n]{x+1}-\sqrt [n] x\right)-1$](http://latex.artofproblemsolving.com/e/f/b/efbe0b5f22cefaddefc75fc54fe3f8d3ccd4b7ea.png)





where



















![$f : [a,b]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/8/c/8/8c8b7678e319f142f7a20ab65d6a4b798011f2d7.png)


![$x_n\in [a,b]$](http://latex.artofproblemsolving.com/1/5/d/15d01c2b442eb61f8c5dac03a2ac5acddd17ee99.png)


![$f: [a,b]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/2/3/3/2334ef1734f81e546c098e61219efa02cb4104e8.png)


Prove that exists

![$(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$](http://latex.artofproblemsolving.com/d/2/9/d292f5cecdee2a67e4753a7473148bdcaf2a9b9c.png)

![$f: [a,b]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/2/3/3/2334ef1734f81e546c098e61219efa02cb4104e8.png)



Prove that exists

![$m(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=n(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right]$](http://latex.artofproblemsolving.com/d/e/2/de2dbf02cc1c975ed18e809764f40f691e38308e.png)







where






![$f(x)=\sqrt [3]{x^3+ax^2+1}+\sqrt [3]{x^3-ax^2+1}-2x$](http://latex.artofproblemsolving.com/c/3/4/c34987db6fe60b296432e9153c04130da60ade69.png)


Prove that exists






![$f:[a,b]\rightarrow\mathbb R$](http://latex.artofproblemsolving.com/2/1/2/212b2745cf32bdb4ace7d1a8510395ecc6aee0aa.png)



Lagrange's theorem is uniquely and


Proof. The point




Denote





Can obtain the inequalities











Proof. Since





for any


![$[\sin x,x]$](http://latex.artofproblemsolving.com/4/a/b/4ab3a58a23ed3e468cf24365396116972bbb7588.png)
![$[x,\tan x]$](http://latex.artofproblemsolving.com/2/3/e/23ed8463c02cf2ca6696bf0513db063376787378.png)
Therefore, exist












![$P\in \mathbb R[X]$](http://latex.artofproblemsolving.com/d/8/e/d8e203cf820f9c3adcda5e6db0a92598d090fd80.png)




Proof. If



![$f:[a,b]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/0/b/b/0bb1e0310b7cbff4a1d3f47a70e6925fca11a9ca.png)

Observe that





![$f'(x)=e^{\frac xk}\cdot\left[P'(x)+\frac 1k\cdot P(x)\right]$](http://latex.artofproblemsolving.com/2/0/2/202226d9e573c4a3bce45b7bcd6cc2540be98ad3.png)





Proof. Consider
![$f:[k,k+1]\rightarrow\mathbb R$](http://latex.artofproblemsolving.com/8/9/f/89f69111693efe083a26a487c3a9047093e532f7.png)



Lagrange's theorem results that exists
![$c_k\in [k,k+1]$](http://latex.artofproblemsolving.com/8/f/1/8f1396a42931be2b50e4503181377a25ef21cca3.png)







![$s_n\nearrow L\in [2,3]$](http://latex.artofproblemsolving.com/7/7/1/771f47ff986e6d0371784358d3d44093c5fd6af4.png)
PP15. For a given





Proof. Since
















![$\sqrt [3]{\cot a}-\tan x\ .s.s.$](http://latex.artofproblemsolving.com/f/8/4/f841aadc8b1a353e8f0e19ebbf1504b825599614.png)
![$\arctan\left(\sqrt [3]{\cot a}\right)-x\implies$](http://latex.artofproblemsolving.com/2/1/2/2129bbdd9fff8d0146e23c251a6bef2f26347c8c.png)
![$f'(x)\ .s.s.\ \arctan\left(\sqrt [3]{\cot a}\right)-x$](http://latex.artofproblemsolving.com/6/2/9/629df0daab0eca60f017ded9e4ca89bf1633a959.png)
In conclusion, the function


![$\left(0,\arctan\left(\sqrt [3]{\cot a}\right)\right]$](http://latex.artofproblemsolving.com/8/1/e/81eae77ee8f4f0afd885962493f5ed57c5f976c2.png)

![$\left[\arctan\left(\sqrt [3]{\cot a}\right),\frac {\pi}{2}\right)\implies$](http://latex.artofproblemsolving.com/2/3/f/23f89fe267e40b234904702ae1431034f3ae704d.png)
![$\boxed{\ \max_{x\in \left(0,\frac {\pi}{2}\right)}f(x)=f\left[\arctan\left(\sqrt [3]{\cot a}\right)\right]\ }$](http://latex.artofproblemsolving.com/5/7/a/57a490d74c8e278a325e58ee350076ad237e75bd.png)
PP16. Find


Proof. Let














![$x\in [1,e]\implies$](http://latex.artofproblemsolving.com/5/2/d/52d7e03b5f7203d24196c628a7f1b5eebc129e6b.png)





we have
![$I'(a)=\left(-\frac {2}{a^2}+\frac {e+1}{a}\right)\sim \left[(e+1)a-2\right]\sim \left(a-\frac {2}{e+1}\right)$](http://latex.artofproblemsolving.com/0/5/0/050d6aa725cf99539f57c41268113ecc6b5cb67c.png)







In conclusion,


Therefore, the evolution of the function



See here and here
This post has been edited 28 times. Last edited by Virgil Nicula, Nov 26, 2015, 3:45 PM