Finding the point of differentiability of a real valued function.

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Let $f:(0,2)rightarrow mathbbR$ be defined by

$f(x)=x^2 $ if $xin (0,2)cap mathbbQ$ and

$f(x)=2x-1 $ if $xin (0,2)- mathbbQ$

Check for the points of differentiability of $f$ and evaluate the derivative at those points.

My attempt: I know that this function $f$ is only in continuous only at the point $x=1$. But I am not sure how to check the differentiability at the point $x=1$. Any suggestions are welcome.

real-analysis derivatives continuity

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asked Jul 23 at 8:29

Babai

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up vote
0
down vote

favorite

Let $f:(0,2)rightarrow mathbbR$ be defined by

$f(x)=x^2 $ if $xin (0,2)cap mathbbQ$ and

$f(x)=2x-1 $ if $xin (0,2)- mathbbQ$

Check for the points of differentiability of $f$ and evaluate the derivative at those points.

My attempt: I know that this function $f$ is only in continuous only at the point $x=1$. But I am not sure how to check the differentiability at the point $x=1$. Any suggestions are welcome.

real-analysis derivatives continuity

share|cite|improve this question

asked Jul 23 at 8:29

Babai

2,50021539

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $f:(0,2)rightarrow mathbbR$ be defined by

$f(x)=x^2 $ if $xin (0,2)cap mathbbQ$ and

$f(x)=2x-1 $ if $xin (0,2)- mathbbQ$

Check for the points of differentiability of $f$ and evaluate the derivative at those points.

My attempt: I know that this function $f$ is only in continuous only at the point $x=1$. But I am not sure how to check the differentiability at the point $x=1$. Any suggestions are welcome.

real-analysis derivatives continuity

share|cite|improve this question

asked Jul 23 at 8:29

Babai

2,50021539

Let $f:(0,2)rightarrow mathbbR$ be defined by

$f(x)=x^2 $ if $xin (0,2)cap mathbbQ$ and

$f(x)=2x-1 $ if $xin (0,2)- mathbbQ$

Check for the points of differentiability of $f$ and evaluate the derivative at those points.

My attempt: I know that this function $f$ is only in continuous only at the point $x=1$. But I am not sure how to check the differentiability at the point $x=1$. Any suggestions are welcome.

real-analysis derivatives continuity

share|cite|improve this question

asked Jul 23 at 8:29

Babai

2,50021539

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 8:29

Babai

2,50021539

asked Jul 23 at 8:29

Babai

2,50021539

asked Jul 23 at 8:29

Babai

2,50021539

2,50021539

add a comment | 

add a comment | 

1 Answer
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2
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Your function isn't continues on $(0,2)$ unless in $x=1$. For differentiability in this point by definition
$$f'(1)=lim_xto1dfracf(x)-f(1)x-1=
begincases
lim_xto1dfracx^2-1x-1=2&xinmathbbQ\
lim_xto1dfrac2x-1-1x-1=2&xnotinmathbbQ
endcases
$$
so the function is continues and differentiable in $x=1$.

share|cite|improve this answer

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
  – Babai
  Jul 23 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Babai Fixed. !
  – Nosrati
  Jul 23 at 11:16
  

  
  
  
  

add a comment | 

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Your function isn't continues on $(0,2)$ unless in $x=1$. For differentiability in this point by definition
$$f'(1)=lim_xto1dfracf(x)-f(1)x-1=
begincases
lim_xto1dfracx^2-1x-1=2&xinmathbbQ\
lim_xto1dfrac2x-1-1x-1=2&xnotinmathbbQ
endcases
$$
so the function is continues and differentiable in $x=1$.

share|cite|improve this answer

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
  – Babai
  Jul 23 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Babai Fixed. !
  – Nosrati
  Jul 23 at 11:16
  

  
  
  
  

add a comment | 

up vote
2
down vote

Your function isn't continues on $(0,2)$ unless in $x=1$. For differentiability in this point by definition
$$f'(1)=lim_xto1dfracf(x)-f(1)x-1=
begincases
lim_xto1dfracx^2-1x-1=2&xinmathbbQ\
lim_xto1dfrac2x-1-1x-1=2&xnotinmathbbQ
endcases
$$
so the function is continues and differentiable in $x=1$.

share|cite|improve this answer

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
  – Babai
  Jul 23 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Babai Fixed. !
  – Nosrati
  Jul 23 at 11:16
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Your function isn't continues on $(0,2)$ unless in $x=1$. For differentiability in this point by definition
$$f'(1)=lim_xto1dfracf(x)-f(1)x-1=
begincases
lim_xto1dfracx^2-1x-1=2&xinmathbbQ\
lim_xto1dfrac2x-1-1x-1=2&xnotinmathbbQ
endcases
$$
so the function is continues and differentiable in $x=1$.

share|cite|improve this answer

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

Your function isn't continues on $(0,2)$ unless in $x=1$. For differentiability in this point by definition
$$f'(1)=lim_xto1dfracf(x)-f(1)x-1=
begincases
lim_xto1dfracx^2-1x-1=2&xinmathbbQ\
lim_xto1dfrac2x-1-1x-1=2&xnotinmathbbQ
endcases
$$
so the function is continues and differentiable in $x=1$.

share|cite|improve this answer

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 11:16

edited Jul 23 at 11:16

edited Jul 23 at 11:16

answered Jul 23 at 8:37

Nosrati

19.4k41544

answered Jul 23 at 8:37

Nosrati

19.4k41544

answered Jul 23 at 8:37

Nosrati

19.4k41544

19.4k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
  – Babai
  Jul 23 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Babai Fixed. !
  – Nosrati
  Jul 23 at 11:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
  – Babai
  Jul 23 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Babai Fixed. !
  – Nosrati
  Jul 23 at 11:16
  

  
  
  
  

When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
– Babai
Jul 23 at 8:50

When u compute the right-hand limit, why are considering $x'$s only rational numbers? $x$ can also be an irrational number. But I guess in any case the limit will be 2. Is not it?
– Babai
Jul 23 at 8:50

@Babai Fixed. !
– Nosrati
Jul 23 at 11:16

@Babai Fixed. !
– Nosrati
Jul 23 at 11:16

add a comment | 

 

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