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The question is wrong . I will prove it .
Put x,y=0f(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 getf(0)=1..or..f′(0)=0
Now put x=1 in (1)
we get y2f(y)−2=f′(y)...(2)
Now let f(0)=1 , we have f(1)=2then 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=xt
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.
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(since f(0)=0)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)
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Top NewestThe 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 getf(0)=1..or..f′(0)=0 Now put x=1 in (1) we get y2f(y)−2=f′(y)...(2) Now let f(0)=1 , we have f(1)=2then 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.
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Hey, if you see the other post, which is the same question, here,
you'll see we get,
f(x)−1=xt
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.
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f(1)=2 and f'(0)=0 should be there for the question to be correct
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Ya.. I also reached the same conclusion...
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But your suggestion makes the functional equation pretty much useless, as for x=1, and y=c, c∈dom(f) we do not get any relation to solve for f(x)
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Please if you can post a solution@parv mor
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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)=hf(1+h)−f(1)
⟹f′(1)=hf(1(1+h))−f(1)
From the functional equation we have,
f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))
⟹f(1+h)=f(1)f(1+h)−f(1)−f(1+h)+2
∴f1(x)=hf(1)f(1+h)−f(1)−f(1+h)+2−f(1)=hf(1)(f(1+h)−2)−(f(1+h)−2)
⟹f1(x)=h(f(1)−1)(f(1+h)−2)
⟹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.
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You have to take the limit for the derivative, as h goes to 0.
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Is the question all right?? Put y=0 in the given functional equation to get: 0+2=f(x)+0+0 (since f(0)=0) 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)
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