# Help with a functions question (1 Viewer)

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#### gaganpreetsingh

Determine all functions f(x) defined for all real numbers x and taking real numbers as their values such that
f(x) + f(1-x) = x^2 - 1

If i replace x by 1-x i get the same LHS and if i equate them i get x=0.5. What I need to do more is what i don't know.

PS sorry if i used the wrong group.

#### Hurkyl

Staff Emeritus
Gold Member
Is f supposed to be continuous? Differentiable?

Anyways, one thing I often do with these types of problems is to figure out if I know what any of its values have to be. Then, I'll see how many values of the function I can express in terms of one particular point. E.G. I'll try to express everything I can in terms of, say, f(0), or maybe f(a) (where I let a be an unspecified constant)

#### gaganpreetsingh

I copied the question as it is.

#### AKG

Homework Helper
Something seems wrong:

0 = 1² - 1 = f(1) + f(1-1) = f(1) + f(0) = f(0) + f(1) = f(0) + f(1-0) = 0² - 1 = -1

so 0 = -1. In general, for all real r:

2r
= (r² + r - 0.75) - (r² - r - 0.75)
= [(0.5 + r)² - 1] - [(0.5 - r)² - 1]
= [f(0.5 + r) + f(1 - (0.5 + r))] - [f(0.5 - r) + f(1 - (0.5 - r))]
= [f(0.5 + r) + f(0.5 - r)] - [f(0.5 - r) + f(0.5 + r)]
= f(0.5 + r) + f(0.5 - r) - f(0.5 - r) - f(0.5 + r)
= 0

so for all real r, 2r = 0, i.e. r = 0. This is clearly a contradiction (not every real number is zero) so there are no functions f whose domain is all of the reals and satisfy, for each x in the reals:

f(x) + f(1-x) = x² - 1

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