This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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

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

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

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.

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)

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*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:

TopNewestThe 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

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)

Log in to reply