Find all real functions f from R →R satisfying the relationf(x2 + y

1.	Find all real functions f from R →R satisfying the relationf(x2 + yf(x)) = xf(x + y):
Answer» Put x = 0 and we get f(yf(0)) = 0. If f(0) ≠ 0, then yf(0) takes all real values when y varies over real line. We get f(x) ≡ 0. Suppose f(0) = 0. Taking y = -x, we get f(x² - xf(x)) = 0 for all real x. Suppose there exists x₀ ≠ 0 in R such that f(x₀) = 0. Putting x = x₀ in the given relation we get f(x²₀) = x₀f(x₀ + y); for all y ∈ R. Now the left side is a constant and hence it follows that f is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where f(x) ≠ 0 for all x ≠ 0. Since f(x² - xf(x)) = 0, we conclude that x² - xf(x) = 0 for all x ≠ 0. This implies that f(x) = x for all x ≠ 0. Since f(0) = 0, we conclude that f(x) = x for all x ∈ R. Thus we have two functions: f(x) ≡ 0 and f(x) = x for all x ∈ R.

Find all real functions f from R →R satisfying the relationf(x2 + yf(x)) = xf(x + y):

Answer»

Put x = 0 and we get f(yf(0)) = 0. If f(0) ≠ 0, then yf(0) takes all real values when y varies over real line. We get f(x) ≡ 0. Suppose f(0) = 0. Taking y = -x, we get f(x² - xf(x)) = 0 for all real x.

Suppose there exists x₀ ≠ 0 in R such that f(x₀) = 0. Putting x = x₀ in the given relation we get

f(x²₀) = x₀f(x₀ + y);

for all y ∈ R. Now the left side is a constant and hence it follows that f is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where f(x) ≠ 0 for all x ≠ 0.

Since f(x² - xf(x)) = 0, we conclude that x² - xf(x) = 0 for all x ≠ 0. This implies that f(x) = x for all x ≠ 0. Since f(0) = 0, we conclude that f(x) = x for all x ∈ R.

Thus we have two functions: f(x) ≡ 0 and f(x) = x for all x ∈ R.

Find all real functions f from R →R satisfying the relationf(x2 + yf(x)) = xf(x + y):

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