AoPS Academy
![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
. If 
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, then show that 
for all 
.
over all triples 
of distinct real numbers such that
such that 
.
then using Mean Value Theorem, there exist an such that 
case, suppose 
and assume 
on the whole interval 
. Then by the mean value theorem, there is some such that 
so 
. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of 
. If 
and , that forces 
on the interval ![$[0,1/2]$](http://latex.artofproblemsolving.com/8/2/4/8240ac9beaadf3f2654c3ba09a3987a7305539c5.png)
. Moving the graph up or down at any point makes the slope larger than 
somewhere. Similarly, 
on the interval ![$[1/2,1]$](http://latex.artofproblemsolving.com/5/5/8/5581510e950d8ceb9bcbcfd0dc988d9c4538c684.png)
. for every ![$x\in [0,1/2]$](http://latex.artofproblemsolving.com/0/f/d/0fd16eeab1cb0c53e329c87cea809db55ea5f207.png)
. On any interval ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
with 
, we have 
for some 
, depending on 
. Since we're assuming 
is always bounded by , this means that 
so 
for all ![$a\in[0,1/2]$](http://latex.artofproblemsolving.com/e/0/c/e0c5302f45677b07a169c44aa1085cf1dace8463.png)
.
, then we could use the mean value theorem on the interval ![$[a,1/2]$](http://latex.artofproblemsolving.com/f/7/c/f7c796537cc023b8e3d0dd550cf1c3bdfc0278d4.png)
to get 
But the left side is larger than , and the right side is at most , a contradiction. on the interval , so 
However, this function is not differentiable at 
, so we are done.
:
Assume towards a contradiction that 
.
and 
:![\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]](http://latex.artofproblemsolving.com/4/6/d/46d07da0b23e15de91ca1912093a406842150f21.png)
LMVT on and 
:![\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]](http://latex.artofproblemsolving.com/9/2/c/92c82c921fad2971d842ec2dba87a5e4725c3e97.png)
Differentiability at 
implies 
is not possible.
and 
, reflecting about the line if necessary, we may assume 
.
and 
:![\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]](http://latex.artofproblemsolving.com/8/3/5/83584e1961403c7cf5c6891281a4cbb97346d193.png)
continuity at 
:![\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]](http://latex.artofproblemsolving.com/9/d/a/9da78481512b68b5c3e9ed771a83552d9878e137.png)
contradicts 
for 
.
is differentiable on . For 
and 
you can hopefully do!! put the value of 
in the two equations,So clearly visible that not differentiable on ....
and 
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and 
2. 
3. 
Bracket 2
2. 
3. 
Bracket 3![$$\int_0^{1}\frac{dx}{1-\sqrt[3]{1-\sqrt{x}}}$$](http://latex.artofproblemsolving.com/d/5/e/d5e8db00985944b2eca341e37a94e34661cb5c86.png)
2. 
3. 
Bracket 4
2. 
3. 
Bracket 5
2. 
3. 
Bracket 6
2. 
3. 
Bracket 7
2. 
3. 
Bracket 8
2. 
3. 
![$$\int_0^{2\pi}\sqrt[4]{\cos^4(x)-\cos(2x)}dx$$](http://latex.artofproblemsolving.com/7/e/e/7eebb4a488fec702e0130a253ef87f0f6dfedb2d.png)
2. ![$$\int_0^1 \frac{x^5}{\sqrt[3]{1-x^3}}dx$$](http://latex.artofproblemsolving.com/6/0/5/605bb9d81d0ccff54a276c9f7bf1c1726c136398.png)
3. 
Bracket 2
2. 
3. 
Bracket 3
2. ![$$\int_0^1 \frac{dx}{(\sqrt{x}+\sqrt[3]{x})^2}$$](http://latex.artofproblemsolving.com/6/9/8/69869c0ed754c0dfacf412ac6411742ffa164a6d.png)
3. 
Bracket 4
2. 
3. 
2. 
3. 
Bracket 2
2. 
3. 
2. 
3. 
2. 
3. 
4. 
5. 
be a function which is continuous on ![\( [0,1] \)](http://latex.artofproblemsolving.com/0/1/0/01026be294289be8f090c7b2586dde64a0d5b479.png)
and differentiable on 
,
.
such that 
.
such that 
