perplexus.info :: Just Math : Function Challenge #2

Function Challenge #2 (Posted on 2024-11-16)

Let f be a real valued function defined on all real numbers. Given that:

f(xy + 1) = f(x)f(y) − f(y) − x + 2

for all x, y and f(0) = 1

Find f(x).

No Solution Yet Submitted by K Sengupta

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Solution

Comment 1 of 1

Let (x,y) = (0,0)

Then f(1) = f(0)*f(0) - f(0) - 0 + 2

f(1) = 1*1 - 1 - 0 + 2

f(1) = 2

Now let (x,y) = (x,0)

Then f(1) = f(x)*f(0) - f(0) - x + 2

2 = f(x)*1 - 1 - x + 2

2 = f(x) - x + 1

f(x) = x+1

Check:

f(xy+1) = f(x)f(y) − f(y) − x + 2

(xy+1) + 1 = (x+1)*(y+1) - (y+1) - x + 2

xy + 2 = xy + x + y + 1 - x - y + 1

xy + 2 = xy + 2

Posted by Brian Smith on 2024-11-16 15:37:04

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