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How find this $f(overline{E})=overline{f(E)} Longleftrightarrow ?$True/False Real Analysis Qualifying Exam QuestionsSufficient condition for inflection pointDoes $max{x|h_n(x) = 0} to max{x|h(x) = 0}$ provided $h_n$ is continuous, convex and $h_n to h$ uniformly?Derive an $O(h^4)$ five point formula to approximate $f'(x_0)$ that uses $f(x_0 - h), f(x_0), f(x_0 + h), f(x_0 + 2h), f(x_0 + 3h)$True or false, prove or find a counterexample. If ${x_n}$ is a sequence such that ${x_n^2}$ converges, then ${x_n}$ convergesProve that $f(x)=begin{cases}1& text{if x is rational},\0 &text{if x is irrational}end{cases}$ is discontinuous at every real number.If $f$ is analytic on $(a, b)$ containing at point $x_{0}$ with $f^{(n)}(x_{0}) = 0$ for $n in mathbb{N}$, prove $f(x) = 0$ for all $x$.Writing proofs to show that a set is open, not closed and infinite

1

$begingroup$

Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.

Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)

real-analysis derivatives

share|cite|improve this question

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:18
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.

Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)

real-analysis derivatives

share|cite|improve this question

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:18
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.

Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)

real-analysis derivatives

share|cite|improve this question

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

$endgroup$

share|cite|improve this question

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

share|cite|improve this question

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

edited Mar 16 at 12:01

daw

24.8k1745

edited Mar 16 at 12:01

daw

24.8k1745

asked Mar 15 at 17:29

JankaJanka

105

asked Mar 15 at 17:29

JankaJanka

105

1 Answer
1

active

oldest

votes

1

$begingroup$

If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$

This limit exists iff $f'(x)$ exists or $f(x)=0$

share|cite|improve this answer

edited Mar 16 at 11:25

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but how did $2f(x)$ appear ? From what ?
  $endgroup$
  – Janka
  Mar 15 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  and what does $h$ mean ?
  $endgroup$
  – Janka
  Mar 15 at 19:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f$ has to be continuous in $x_0$, too.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:16
  

  
  
  
  

 | 
show 2 more comments

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1

$begingroup$

If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$

This limit exists iff $f'(x)$ exists or $f(x)=0$

share|cite|improve this answer

edited Mar 16 at 11:25

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but how did $2f(x)$ appear ? From what ?
  $endgroup$
  – Janka
  Mar 15 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  and what does $h$ mean ?
  $endgroup$
  – Janka
  Mar 15 at 19:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f$ has to be continuous in $x_0$, too.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:16
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$

This limit exists iff $f'(x)$ exists or $f(x)=0$

share|cite|improve this answer

edited Mar 16 at 11:25

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but how did $2f(x)$ appear ? From what ?
  $endgroup$
  – Janka
  Mar 15 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  and what does $h$ mean ?
  $endgroup$
  – Janka
  Mar 15 at 19:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
  $endgroup$
  – Gabriele Cassese
  Mar 15 at 19:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f$ has to be continuous in $x_0$, too.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:16
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$

This limit exists iff $f'(x)$ exists or $f(x)=0$

share|cite|improve this answer

edited Mar 16 at 11:25

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

$endgroup$

share|cite|improve this answer

edited Mar 16 at 11:25

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316

answered Mar 15 at 19:33

Gabriele CasseseGabriele Cassese

1,171316