Is the question all right??
Put y=0 in the given functional equation to get:
\[0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}}\]
\[\Large f(x)=2\]
which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0)

The question is wrong . I will prove it .
Put \(x,y=0\)
\[f(0)^{2}-3f(0)+2=0\]
This implies \[f(0)=1,2\]
now put \(x,y=1\)
this gives \(f(1)=1,2\)
Differentiating the function partially with respect to \(x\) we get
\[f'(x)f(y)=f'(x)+yf'(xy).......(1)\]
Now put \(x,y=1\) in \((1)\)
We get \(f(1)=2\)
Now put \(x,y=0\) in \((1)\)
We get\(f(0)=1..or..f'(0)=0\)
Now put \(x=1\) in (1)
we get \(\frac{2f(y)-2}{y}=f'(y)...(2)\)
Now let \(f(0)=1\) , we have \(f(1)=2\)then we observe that \[f(x)=x+1\]
also in satisfies main functional equation and (2),(1) but \[f'(x)=1\] hence question is wrong .
If only main functional equation is given without any condition then \[f(x)=x+1\].

Hey, if you see the other post, which is the same question, here,
you'll see we get,
\( f(x) - 1 = x^t \)
For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again.

But your suggestion makes the functional equation pretty much useless, as for \(x=1\), and \(y=c\), \(c\in dom(f)\) we do not get any relation to solve for \(f(x)\)

I also came up with the same conclusion as you, i.e. \(f(x)=2\), so i thought of using the \(f'(1)=2\).
So as we know from the definition of derivatives

\(f'(1)=\frac { f(1+h)-f(1) }{ h }\\ \)

\(\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h } \)

\(\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }\)

\(\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h } \)

\(\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)\)

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIs the question all right?? Put y=0 in the given functional equation to get: \[0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}}\] \[\Large f(x)=2\] which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0)

Log in to reply

The question is wrong . I will prove it . Put \(x,y=0\) \[f(0)^{2}-3f(0)+2=0\] This implies \[f(0)=1,2\] now put \(x,y=1\) this gives \(f(1)=1,2\) Differentiating the function partially with respect to \(x\) we get \[f'(x)f(y)=f'(x)+yf'(xy).......(1)\] Now put \(x,y=1\) in \((1)\) We get \(f(1)=2\) Now put \(x,y=0\) in \((1)\) We get\(f(0)=1..or..f'(0)=0\) Now put \(x=1\) in (1) we get \(\frac{2f(y)-2}{y}=f'(y)...(2)\) Now let \(f(0)=1\) , we have \(f(1)=2\)then we observe that \[f(x)=x+1\] also in satisfies main functional equation and (2),(1) but \[f'(x)=1\] hence question is wrong . If only main functional equation is given without any condition then \[f(x)=x+1\].

Log in to reply

Hey, if you see the other post, which is the same question, here,

you'll see we get,

\( f(x) - 1 = x^t \)

For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again.

Log in to reply

f(1)=2 and f'(0)=0 should be there for the question to be correct

Log in to reply

Ya.. I also reached the same conclusion...

Log in to reply

But your suggestion makes the functional equation pretty much useless, as for \(x=1\), and \(y=c\), \(c\in dom(f)\) we do not get any relation to solve for \(f(x)\)

Log in to reply

Please if you can post a solution@parv mor

Log in to reply

I also came up with the same conclusion as you, i.e. \(f(x)=2\), so i thought of using the \(f'(1)=2\). So as we know from the definition of derivatives

\(f'(1)=\frac { f(1+h)-f(1) }{ h }\\ \)

\(\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h } \)

From the functional equation we have,\( \\ \)

\(f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))\)

\(\Longrightarrow f(1+h)=f(1)f(1+h)-f(1)-f(1+h)+2\)

\(\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }\)

\(\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h } \)

\(\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)\)

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong.

Log in to reply

You have to take the limit for the derivative, as h goes to 0.

Log in to reply