Help: Functional Equation

If $f(x)$ is a polynomial in $x$ satisfying $f(x) f(y) + 2 = f(x) + f(y) + f(xy)$ and $f(0) = 0, f'(1) = 2$ , find $f(x)$ .

#Algebra

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:

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
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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)

Rishabh Jain - 5 years, 4 months ago

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.

Akhilesh Prasad - 5 years, 4 months ago

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

Ameya Daigavane - 5 years, 3 months ago

Please if you can post a solution@parv mor

Akhilesh Prasad - 5 years, 4 months ago

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

parv mor - 5 years, 4 months ago

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 $f(x)$

Akhilesh Prasad - 5 years, 4 months ago

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

Rishabh Jain - 5 years, 4 months ago

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.

Ameya Daigavane - 5 years, 3 months ago

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 $f(x)=x+1$ .

Shivam Jadhav - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$