Prove that $f(x)=ax+b$












0














Im stuck in this exercise:



$frac{f(x+h)-f(x)}{h}=f'(x) spacespace forall xland forall h $



Prove that $space f(x)=ax+b$



I think it's something easy but I can not find a concrete proof.



I would appreciate a hint. Thanks






calculus real-analysis







share|cite|improve this question


















  • 2




    $frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
    – fleablood
    Nov 27 at 3:20
















0














Im stuck in this exercise:



$frac{f(x+h)-f(x)}{h}=f'(x) spacespace forall xland forall h $



Prove that $space f(x)=ax+b$



I think it's something easy but I can not find a concrete proof.



I would appreciate a hint. Thanks






calculus real-analysis







share|cite|improve this question


















  • 2




    $frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
    – fleablood
    Nov 27 at 3:20














0












0








0







Im stuck in this exercise:



$frac{f(x+h)-f(x)}{h}=f'(x) spacespace forall xland forall h $



Prove that $space f(x)=ax+b$



I think it's something easy but I can not find a concrete proof.



I would appreciate a hint. Thanks






calculus real-analysis







share|cite|improve this question













Im stuck in this exercise:



$frac{f(x+h)-f(x)}{h}=f'(x) spacespace forall xland forall h $



Prove that $space f(x)=ax+b$



I think it's something easy but I can not find a concrete proof.



I would appreciate a hint. Thanks






calculus real-analysis




calculus real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 at 2:06









Ezequiel Saidman

93




93








  • 2




    $frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
    – fleablood
    Nov 27 at 3:20














  • 2




    $frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
    – fleablood
    Nov 27 at 3:20








2




2




$frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
– fleablood
Nov 27 at 3:20




$frac{f(x+h)-f(x)}{h}=f'(x)$ is not possible for $h=0$.
– fleablood
Nov 27 at 3:20










2 Answers
2






active

oldest

votes


















5














Applying the condition with $x=0$ we have $f(h) = f(0) + h f'(0)$ for all $h$, et voilà.






share|cite|improve this answer























  • Thanks! it's easier than i expected.
    – Ezequiel Saidman
    Nov 27 at 2:19








  • 1




    What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
    – fleablood
    Nov 27 at 3:17










  • @fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
    – angryavian
    Nov 27 at 4:24










  • @fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
    – Paramanand Singh
    Nov 27 at 4:49



















1














I am elaborating @angryavian 's answer.



When $x=0$, observe that



$$ frac{f(h)-f(0)}{h}= f'(0) iff f(h)=f'(0)h+f(0) hspace{4mm} forall h$$



Thus, by switching the letter from $h$ to $x$,



$$f(x)=f'(0)x+f(0) $$



where $a=f'(0)$ and $b=f(0)$ in this case.






share|cite|improve this answer























  • You haven't said what $y$ is.
    – Shaun
    Nov 27 at 2:19










  • I think it refers to y=h
    – Ezequiel Saidman
    Nov 27 at 2:21












  • @Shaun Thanks, I edited
    – LeB
    Nov 27 at 2:22










  • You're welcome. That's better :)
    – Shaun
    Nov 27 at 2:22






  • 1




    As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
    – fleablood
    Nov 27 at 3:25











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














Applying the condition with $x=0$ we have $f(h) = f(0) + h f'(0)$ for all $h$, et voilà.






share|cite|improve this answer























  • Thanks! it's easier than i expected.
    – Ezequiel Saidman
    Nov 27 at 2:19








  • 1




    What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
    – fleablood
    Nov 27 at 3:17










  • @fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
    – angryavian
    Nov 27 at 4:24










  • @fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
    – Paramanand Singh
    Nov 27 at 4:49
















5














Applying the condition with $x=0$ we have $f(h) = f(0) + h f'(0)$ for all $h$, et voilà.






share|cite|improve this answer























  • Thanks! it's easier than i expected.
    – Ezequiel Saidman
    Nov 27 at 2:19








  • 1




    What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
    – fleablood
    Nov 27 at 3:17










  • @fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
    – angryavian
    Nov 27 at 4:24










  • @fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
    – Paramanand Singh
    Nov 27 at 4:49














5












5








5






Applying the condition with $x=0$ we have $f(h) = f(0) + h f'(0)$ for all $h$, et voilà.






share|cite|improve this answer














Applying the condition with $x=0$ we have $f(h) = f(0) + h f'(0)$ for all $h$, et voilà.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 2:20









Shaun

8,647113680




8,647113680










answered Nov 27 at 2:10









angryavian

38.8k23180




38.8k23180












  • Thanks! it's easier than i expected.
    – Ezequiel Saidman
    Nov 27 at 2:19








  • 1




    What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
    – fleablood
    Nov 27 at 3:17










  • @fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
    – angryavian
    Nov 27 at 4:24










  • @fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
    – Paramanand Singh
    Nov 27 at 4:49


















  • Thanks! it's easier than i expected.
    – Ezequiel Saidman
    Nov 27 at 2:19








  • 1




    What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
    – fleablood
    Nov 27 at 3:17










  • @fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
    – angryavian
    Nov 27 at 4:24










  • @fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
    – Paramanand Singh
    Nov 27 at 4:49
















Thanks! it's easier than i expected.
– Ezequiel Saidman
Nov 27 at 2:19






Thanks! it's easier than i expected.
– Ezequiel Saidman
Nov 27 at 2:19






1




1




What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
– fleablood
Nov 27 at 3:17




What if $h = 0$? Then you have $frac {f(x)-f(x)}0 = frac 00 = f'(x)$ and that's not defined.
– fleablood
Nov 27 at 3:17












@fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
– angryavian
Nov 27 at 4:24




@fleablood When $h ne 0$, the assumption implies $f(h) = f(0) + h f'(0)$. When $h = 0$, we automatically have $f(h) = f(0) + h f'(0)$.
– angryavian
Nov 27 at 4:24












@fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
– Paramanand Singh
Nov 27 at 4:49




@fleablood : I think the issue is with the question. It needs to say that the condition holds for all $x$ and all non-zero $h$. And then the relation $f(h) =f(0)+hf'(0)$ holds for all $h$ as in this answer.
– Paramanand Singh
Nov 27 at 4:49











1














I am elaborating @angryavian 's answer.



When $x=0$, observe that



$$ frac{f(h)-f(0)}{h}= f'(0) iff f(h)=f'(0)h+f(0) hspace{4mm} forall h$$



Thus, by switching the letter from $h$ to $x$,



$$f(x)=f'(0)x+f(0) $$



where $a=f'(0)$ and $b=f(0)$ in this case.






share|cite|improve this answer























  • You haven't said what $y$ is.
    – Shaun
    Nov 27 at 2:19










  • I think it refers to y=h
    – Ezequiel Saidman
    Nov 27 at 2:21












  • @Shaun Thanks, I edited
    – LeB
    Nov 27 at 2:22










  • You're welcome. That's better :)
    – Shaun
    Nov 27 at 2:22






  • 1




    As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
    – fleablood
    Nov 27 at 3:25
















1














I am elaborating @angryavian 's answer.



When $x=0$, observe that



$$ frac{f(h)-f(0)}{h}= f'(0) iff f(h)=f'(0)h+f(0) hspace{4mm} forall h$$



Thus, by switching the letter from $h$ to $x$,



$$f(x)=f'(0)x+f(0) $$



where $a=f'(0)$ and $b=f(0)$ in this case.






share|cite|improve this answer























  • You haven't said what $y$ is.
    – Shaun
    Nov 27 at 2:19










  • I think it refers to y=h
    – Ezequiel Saidman
    Nov 27 at 2:21












  • @Shaun Thanks, I edited
    – LeB
    Nov 27 at 2:22










  • You're welcome. That's better :)
    – Shaun
    Nov 27 at 2:22






  • 1




    As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
    – fleablood
    Nov 27 at 3:25














1












1








1






I am elaborating @angryavian 's answer.



When $x=0$, observe that



$$ frac{f(h)-f(0)}{h}= f'(0) iff f(h)=f'(0)h+f(0) hspace{4mm} forall h$$



Thus, by switching the letter from $h$ to $x$,



$$f(x)=f'(0)x+f(0) $$



where $a=f'(0)$ and $b=f(0)$ in this case.






share|cite|improve this answer














I am elaborating @angryavian 's answer.



When $x=0$, observe that



$$ frac{f(h)-f(0)}{h}= f'(0) iff f(h)=f'(0)h+f(0) hspace{4mm} forall h$$



Thus, by switching the letter from $h$ to $x$,



$$f(x)=f'(0)x+f(0) $$



where $a=f'(0)$ and $b=f(0)$ in this case.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 at 3:28

























answered Nov 27 at 2:15









LeB

991217




991217












  • You haven't said what $y$ is.
    – Shaun
    Nov 27 at 2:19










  • I think it refers to y=h
    – Ezequiel Saidman
    Nov 27 at 2:21












  • @Shaun Thanks, I edited
    – LeB
    Nov 27 at 2:22










  • You're welcome. That's better :)
    – Shaun
    Nov 27 at 2:22






  • 1




    As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
    – fleablood
    Nov 27 at 3:25


















  • You haven't said what $y$ is.
    – Shaun
    Nov 27 at 2:19










  • I think it refers to y=h
    – Ezequiel Saidman
    Nov 27 at 2:21












  • @Shaun Thanks, I edited
    – LeB
    Nov 27 at 2:22










  • You're welcome. That's better :)
    – Shaun
    Nov 27 at 2:22






  • 1




    As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
    – fleablood
    Nov 27 at 3:25
















You haven't said what $y$ is.
– Shaun
Nov 27 at 2:19




You haven't said what $y$ is.
– Shaun
Nov 27 at 2:19












I think it refers to y=h
– Ezequiel Saidman
Nov 27 at 2:21






I think it refers to y=h
– Ezequiel Saidman
Nov 27 at 2:21














@Shaun Thanks, I edited
– LeB
Nov 27 at 2:22




@Shaun Thanks, I edited
– LeB
Nov 27 at 2:22












You're welcome. That's better :)
– Shaun
Nov 27 at 2:22




You're welcome. That's better :)
– Shaun
Nov 27 at 2:22




1




1




As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
– fleablood
Nov 27 at 3:25




As far as nits go that is a pretty humongous one. It's fairly important that you can't divide by $0$.
– fleablood
Nov 27 at 3:25


















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