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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Japanese Olympiad
parkjungmin   2
N 36 minutes ago by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
2 replies
parkjungmin
Yesterday at 6:51 PM
parkjungmin
36 minutes ago
Japanese high school Olympiad.
parkjungmin   0
38 minutes ago
It's about the Japanese high school Olympiad.

If there are any students who are good at math, try solving it.
0 replies
parkjungmin
38 minutes ago
0 replies
Integration Bee Kaizo
Calcul8er   60
N an hour ago by Svyatoslav
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
60 replies
Calcul8er
Mar 2, 2025
Svyatoslav
an hour ago
Marginal Profit
NC4723   2
N an hour ago by Juno_34
Please help me solve this
2 replies
NC4723
Dec 11, 2015
Juno_34
an hour ago
No more topics!
f'(1)>1 implies f has a fixed point in (0,1)
Sayan   13
N Apr 27, 2025 by Apple_maths60
Source: ISI(BS) 2010 #4
A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.
13 replies
Sayan
May 17, 2012
Apple_maths60
Apr 27, 2025
f'(1)>1 implies f has a fixed point in (0,1)
G H J
Source: ISI(BS) 2010 #4
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Sayan
2130 posts
#1 • 2 Y
Y by Adventure10, Mango247
A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.
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JoeBlow
3469 posts
#2 • 5 Y
Y by Abhinandan18, ring_r, Severus, Adventure10, and 1 other user
Define $g(x)=f(x)-x$. Since $g'(1)>0$ and $g(1)=0$ there is a $0<\xi <1$ such that $g(\xi )<0$. But $g(0)>0$, so by the intermediate value theorem there is an $x_0\in (0,\xi)\subset (0,1)$ such that $g(x_0)=0\Leftrightarrow f(x_0)=x_0$.
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FOURRIER
727 posts
#3 • 6 Y
Y by chipgiboy, ring_r, integrated_JRC, Jaggoseni, Adventure10, Mango247
JoeBlow wrote:
Define $g(x)=f(x)-x$. Since $g'(1)>0$ and $g(1)=0$ there is a $0<\xi <1$ such that $g(\xi )<0$
maybe this little thing have to be clarified:
$0< g'(1) = \lim_{x \to 1^{-}} \frac{g(x)}{x-1}$
then there exists $c$ in $]0,1[$ such that $\frac{g(x)}{x-1} >0 $ $  \forall x \in ]c,1[$
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Aranya
35 posts
#4 • 1 Y
Y by Adventure10
My solution: By the Mean Value Theorem, $\dfrac {f(1)-f(0)}{1-0}=f'(c)$ for some c belonging to (0,1).

That gives $f'(c)<1$ for some c in (0,1).

Now, let $f(x)>x$ for all x in (0,1). Then, $f'(x)>1$ for all x in (0,1) which contradicts our result.

Taking $f(x)<x$ for all x belonging to (0,1) implies $f'(x)<1$ for all x lying in (0,1) which is false by hypothesis.

Therefore, there must be at least one $x_0$ so that $f(x_0)=x_0$ for $x_0$ belonging to (0,1). QED.
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kd7
285 posts
#5 • 1 Y
Y by Adventure10
Aranya wrote:
My solution: By the Mean Value Theorem, $\dfrac {f(1)-f(0)}{1-0}=f'(c)$ for some c belonging to (0,1).

That gives $f'(c)<1$ for some c in (0,1).

Now, let $f(x)>x$ for all x in (0,1). Then, $f'(x)>1$ for all x in (0,1) which contradicts our result.

Taking $f(x)<x$ for all x belonging to (0,1) implies $f'(x)<1$ for all x lying in (0,1) which is false by hypothesis.

Therefore, there must be at least one $x_0$ so that $f(x_0)=x_0$ for $x_0$ belonging to (0,1). QED.

No. Not necessary if $f(x)>x$ that implies $f'(x)>1$
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stranger_02
337 posts
#6
Y by
Well, an alternate solution.

Just consider the line $y=x$.

We claim that $f(x)$ intersects this line at some $x=\beta$, where, $\beta \in (0,1)$

Proof:

We know that the slope of $y=x$ at $x=1$ is $1$.

Given that $f(1)=1$ and $f'(1)>1$. This implies that at the point $1$, $f(x)$ intersects $y=x$ from under the line to above.

Converting this into formal language we may state that $f(1-\epsilon)<1$ and $f(1+\epsilon)>1$ for some $\epsilon \rightarrow 0^+ $

Therefore, when $x \rightarrow 1^-$, $f(x)$ lies below $y=x$ ......... $(\zeta_1)$

Now, given that $f(0)>0$. This clearly states that $f(x)$ lies above $y=x$ when $x=0$ ..........$(\zeta_2)$

Combining $\zeta_1$ and $\zeta_2$ we may conclude that $f(x)$ must have intersected the line $y=x$ at some point in the interval of $(0,1)$.

This proves our claim. $\blacksquare$

$\boxed{\therefore f(\beta)=\beta \text{ for some } \beta \in (0,1)}$ , which is just what was to be proved.

Q.E.D. $\square$
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Maths_1729
390 posts
#7
Y by
Sayan wrote:
A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.

Can Someone Please check This...

Now $f(0)>0, f(1)=1\implies$ By LMVT There exist a constant $k\in(0,1)$ such that $\frac{f(1) -f(0)}{1-0}=f'(k)\implies$

either $1>f'(k)\geq 0$ or $f'(k)<0$
FTSOC Assume that doesn't exist a fixed point $\in (0, 1)$ Then we again by LMVT We can't have any $x_1\in (0, 1)$ Such that $f'(x_1)=1$ But this is never possible As there exist a $k\in (0, 1)$ such that either $f'(k)<0$ or $1>f'(k)\geq 0$ and $f'(1)>1,f(x)$ is continuous So there will always exist a $x_1\in (0, 1)$ such that $f(x_1)=1$ Hence Contradiction.
This post has been edited 1 time. Last edited by Maths_1729, Apr 20, 2021, 8:29 PM
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MathsSolver007
542 posts
#8
Y by
Storage(same as #2)
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Agsh2005
70 posts
#9
Y by
MathsSolver007 wrote:
Storage(same as #2)

This one is a worth mentioning fact(an often overlooked one)

$\textbf{Fact:}$ $\text{For a function f, if}~ \nolinebreak \exists c \text{ such that }  f'(c)>0(\text{or} < 0)  \text{ then f is not necessarily increasing(or decreasing) on any neighbourhood 
around }  c $

Take $$ f(x) = \begin{cases} \frac{x}{2} + x^2 sin(\frac{1}{x}) ~&\text{if} ~ x \neq 0 \\ 0 ~ &\text{if} ~ x = 0 \end{cases}$$
One can check that $f'(0) >0$ but the function is not increasing on any open interval containing 0.

@above your solution is therefore incorrect
This post has been edited 1 time. Last edited by Agsh2005, Jan 30, 2023, 6:56 PM
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Agsh2005
70 posts
#10
Y by
A proper proof for this problem would be the following.

Consider the function $ g(x) = f(x) -x $ on the interval $[0,1]$. By the Extreme Value theorem we can say that $\exists ~ m \in [0,1]$ such that
$ g(m) \leq g(x) ~ \forall x \in [0,1] $ and $ \exists M \in [0,1]$ such that $ g(M)\geq g(x)~ \forall x \in [0,1]$ Now we have that $ g(1)=0$ which gives us $ g(m)\leq 0$ Since $g(0)>0$ we must have that $ M \neq 1$.

$\text{Claim:}$ $m \in (0,1)$.
$\text{Proof:} $ Let us assume FTSoC that $m=1$ .
But that would give us $g'(1)= \lim_{x \to 1^{-}} \frac{g(x)- g(1)} {x-1} < 0 $ which is a contradiction.

We can now conclude by the Intermediate Value Theorem.
This post has been edited 1 time. Last edited by Agsh2005, Jan 30, 2023, 7:28 PM
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MathsSolver007
542 posts
#11
Y by
Agsh2005 wrote:
A proper proof for this problem would be the following.

Consider the function $ g(x) = f(x) -x $ on the interval $[0,1]$. By the Extreme Value theorem we can say that $\exists ~ m \in [0,1]$ such that
$ g(m) \leq g(x) ~ \forall x \in [0,1] $ and $ \exists M \in [0,1]$ such that $ g(M)\geq g(x)~ \forall x \in [0,1]$ Now we have that $ g(1)=0$ which gives us $ g(m)\leq 0$ Since $g(0)>0$ we must have that $ M \neq 1$.

$\text{Claim:}$ $m \in (0,1)$.
$\text{Proof:} $ Let us assume FTSoC that $m=1$ .
But that would give us $g'(1)= \lim_{x \to 1^{-}} \frac{g(x)- g(1)} {x-1} < 0 $ which is a contradiction.

We can now conclude by the Intermediate Value Theorem.

Agreed that my wording was a little bit unclear.

I didn't mean to say that $f$ is increasing or decreasing in the $\varepsilon$ nbd.

The definition of the derivative (the left hand limit) gives us that $\frac{g(1)-g(h)}{1-h} >0$ which in turn implies the existence of some $h$ such that $g(h)>0$. We finish by IVT
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lifeismathematics
1188 posts
#13 • 1 Y
Y by MathSigma
we define $g(x)=f(x)-x$

we have $g(0)>0 , g(1)=0$ and $g'(1)>0$

we propose a $\textbf{\textcolor{red}{Lemma}}:-$ if $h'(c)>0$ , then $\exists$ a $\delta>0$ s.t. $h(x)<h(c)$ for $x \in (c-\delta,c)$

we use the lemma to get for $x^{*} \in (1-\delta,1) , g(x^{*})<g(1)=0$ so we get $g(x^{*})<0$ and $g(0)>0$ by $\textbf{IVT}$ it follows that $\exists$ a $x_{0}$ s.t. $f(x_{0})=x_{0}$ $\blacksquare$
This post has been edited 1 time. Last edited by lifeismathematics, Apr 28, 2023, 10:31 AM
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SatisfiedMagma
458 posts
#14
Y by
Okay, very standard and doable problem. I'm not sure, but prolly irrelevant compared to today's standards? Editing on July 14, 2024: Okay this is not that bad, I should not speak bad about problems :(

Solution: Consider the function $g(x) \coloneqq f(x) - x$. The problem conditions now translate to $g(0) > 0$, $g(1) = 0$ with $g$ differentiable on the interval. We also have that $g'(1) > 0$ with $g$ being continuous since all differentiable functions are continuous. It suffices to show that $g$ has a zero in $(0,1)$.

Lemma: If $g'(1) > 0$, then there is some $\xi$ such that $g(\xi) < 0$ for $\xi \in (0,1)$.

Proof: We will only consider left-hand limit of $g$ at $1$. By the $\epsilon-\delta$ definition of limits, we can conclude that for some $\delta > 0$, we must have
\[1-\delta < x < 1 \implies \left| \frac{g(x)-g(1)}{x-1} -g'(1)\right| < g'(1) \iff -2g'(1) < \frac{g(x)}{1-x} < 0\]which is basically by picking $\epsilon = g'(1)$. Since $1-x > 0$, we can say that there exists there is some $\xi \in (0,1)$ such that $g(\xi) < 0$ which proves the lemma. $\square$
Now note that
\[g'(1) = \lim_{x \to 1}\frac{g(x)-g(1)}{x-1} > 0 \implies \exists \, a \colon g(a) < 0 \text{   for some $a \in (0,1)$.}\]Applying Intermediate Value Theorem on $(0,g(0))$ and$(a,g(a))$, we get that $g(\alpha)$ must be $0$ for some $\alpha \in (0,1)$. This finishes the proof. $\blacksquare$
This post has been edited 1 time. Last edited by SatisfiedMagma, Jul 14, 2024, 11:29 AM
Reason: proving the crucial lemma with "rigor"
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Apple_maths60
26 posts
#15
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g(x) =f(x)-x
Then, g(0)=f(0)>0
g(1)=f(1)-1=0
Since g is continuous and differentiable, we get
g'(x)=f'(x)-1
g'(1)=f'(1)>1, so g is increasing at the point 1
Now we take a k(very small) neighborhood around 1 say (k-1,k+1) the there exist a point c belong to (k-1,k+1) s.t. g(c) <0
So we got g(c)<0 & g(0)>0 , Then by IVT, there exist a point t belong to (0,c) which is subset of (0,1) s.t. g(t)=0
So, g(t)=0
f(t)-t=0
f(t)=t . Done!
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