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Find all real functions f from #RR->RR# satisfying the relation #f(x²+yf(x))=xf(x+y)# ?

Answer 1

Abigail Allen

#f(x)=x#

Making #x=y=0# we have #f(0)=0#

Making #y=0# we have

#f(x^2)=xf(x)# but #f((-x)^2)=f(x^2)=-xf(-x)#

so #f(x)# is an odd function.

Making now #x = -y# we have

#f(x^2-xf(x))=x f(0) = 0#

thus presuming that

#f(x)=0 => x=0#

(remember that #f(x)# is an odd function) we have

#x^2-xf(x)=x(x-f(x))=0#

considering #x ne 0# we have

#f(x)=x#

Verifying

#(x^2+y x)=x(x+y)#

Note. Another way of proof for #f(x)=x# is by doing for #x ge 0#

#f(x) = x^(1/2)f(x^(1/2))= cdots = x^(1/2+1/4+cdots + 1/(2^n))f(x^(1/(2^n)))#

but #lim_(n->oo) x^(1/2^n) = 1# and #lim_(n->oo)(1/2+1/4+cdots+1/(2^n))=1# so

#f(x) = x f(1)# now substituting this result into #f(x^2)=xf(x)# we obtain

#f(x^2)=x^2f(1)->f(1)=1#

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Answer 2

Jace Anderson

The only solutions to the functional equation ( f(x^2 + yf(x)) = xf(x + y) ) are ( f(x) = 0 ) and ( f(x) = x ).

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Answer 3

Savannah Anderson

To find all real functions ( f ) from ( \mathbb{R} ) to ( \mathbb{R} ) satisfying the relation ( f(x^2+yf(x))=xf(x+y) ), we can proceed by substitution and manipulation.

Let ( x = y = 0 ): [ f(0) = 0 ]

Let ( y = 0 ): [ f(x^2) = xf(x) ]

Now, let's substitute ( x = -1 ) in the equation above: [ f(1) = -f(-1) ]

Now, we'll use the equation ( f(x^2) = xf(x) ) and substitute ( x^2 ) with ( x ) in the original equation: [ f(x+f(x)) = xf(x+1) ]

Now, let's substitute ( x = 1 ) in the above equation: [ f(1+f(1)) = f(2) ]

And from the previous deduction, we know ( f(1) = -f(-1) ), so: [ f(-1+f(-1)) = f(0) = 0 ]

This implies ( f(-1) = -1 ).

Now, let's set ( x = 1 ) and ( y = -1 ) in the original equation: [ f(1+(-1)f(1)) = f(0) = 0 ]

So, ( f(0) = 0 ).

Now, using ( f(x^2) = xf(x) ) and ( f(1) = -1 ), we can derive that ( f(x) = x ) for all ( x ) in ( \mathbb{R} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Find all real functions f from #RR-&gt;RR# satisfying the relation #f(x²+yf(x))=xf(x+y)# ?

Find all real functions f from #RR->RR# satisfying the relation #f(x²+yf(x))=xf(x+y)# ?