A derivative problem [on hold]












-6














$f(f(x)+3x+1) = g^2(2x)+x^2$ and
$f'(0) = f'(1) = 4$ then
$g(0) = ?$






functions derivatives







share|cite|improve this question













put on hold as off-topic by Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire Jan 4 at 9:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $g^2(x)=g(g(x))$? In all a case, why it exists?
    – Michael Rozenberg
    Jan 4 at 9:12












  • I dont understand.
    – Hik Aubergine
    Jan 4 at 9:20










  • What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
    – Jimmy Sabater
    Jan 4 at 9:22










  • It is second one
    – Hik Aubergine
    Jan 4 at 9:24










  • Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
    – Jimmy Sabater
    Jan 4 at 9:26
















-6














$f(f(x)+3x+1) = g^2(2x)+x^2$ and
$f'(0) = f'(1) = 4$ then
$g(0) = ?$






functions derivatives







share|cite|improve this question













put on hold as off-topic by Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire Jan 4 at 9:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $g^2(x)=g(g(x))$? In all a case, why it exists?
    – Michael Rozenberg
    Jan 4 at 9:12












  • I dont understand.
    – Hik Aubergine
    Jan 4 at 9:20










  • What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
    – Jimmy Sabater
    Jan 4 at 9:22










  • It is second one
    – Hik Aubergine
    Jan 4 at 9:24










  • Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
    – Jimmy Sabater
    Jan 4 at 9:26














-6












-6








-6


1





$f(f(x)+3x+1) = g^2(2x)+x^2$ and
$f'(0) = f'(1) = 4$ then
$g(0) = ?$






functions derivatives







share|cite|improve this question













$f(f(x)+3x+1) = g^2(2x)+x^2$ and
$f'(0) = f'(1) = 4$ then
$g(0) = ?$






functions derivatives




functions derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 9:08









Hik AubergineHik Aubergine

75




75




put on hold as off-topic by Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire Jan 4 at 9:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire Jan 4 at 9:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, KM101, Shaun, TheSimpliFire

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $g^2(x)=g(g(x))$? In all a case, why it exists?
    – Michael Rozenberg
    Jan 4 at 9:12












  • I dont understand.
    – Hik Aubergine
    Jan 4 at 9:20










  • What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
    – Jimmy Sabater
    Jan 4 at 9:22










  • It is second one
    – Hik Aubergine
    Jan 4 at 9:24










  • Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
    – Jimmy Sabater
    Jan 4 at 9:26


















  • $g^2(x)=g(g(x))$? In all a case, why it exists?
    – Michael Rozenberg
    Jan 4 at 9:12












  • I dont understand.
    – Hik Aubergine
    Jan 4 at 9:20










  • What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
    – Jimmy Sabater
    Jan 4 at 9:22










  • It is second one
    – Hik Aubergine
    Jan 4 at 9:24










  • Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
    – Jimmy Sabater
    Jan 4 at 9:26
















$g^2(x)=g(g(x))$? In all a case, why it exists?
– Michael Rozenberg
Jan 4 at 9:12






$g^2(x)=g(g(x))$? In all a case, why it exists?
– Michael Rozenberg
Jan 4 at 9:12














I dont understand.
– Hik Aubergine
Jan 4 at 9:20




I dont understand.
– Hik Aubergine
Jan 4 at 9:20












What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
– Jimmy Sabater
Jan 4 at 9:22




What do you mean when you write $g^2(2x)$? is the $g ( g (2x) ) $ ? or is it $( g(2x) )^2 $ ?
– Jimmy Sabater
Jan 4 at 9:22












It is second one
– Hik Aubergine
Jan 4 at 9:24




It is second one
– Hik Aubergine
Jan 4 at 9:24












Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
– Jimmy Sabater
Jan 4 at 9:26




Use the chain rule. for example, one has $$ (f'(x) + 3 ) f'( f(x)+3x+1) = 2 g(2x) g' (2x) cdot 2 + 2x $$ Now, plug in $x=0$ and see what happens
– Jimmy Sabater
Jan 4 at 9:26










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