# Functional Equations Resulting from Properties of Functions

## Functional Equations Resulting from Properties of Functions

1. Odd functions having symmetry of graph about the origin: f(x) + f(-x) = 0, $$\forall$$ x ϵ Df.

2. Even functions having symmetry of graph about the y – axis: f(x) = f(-x), $$\forall$$ x ϵ Df.

3. Symmetry of graph about the Point (a, 0): f(a – x) = – f(a + x).

4. Symmetry of graph about the line x = a: f(a – x) = f(a + x).

5. Periodic functions: f(x) = f(x + T) $$\forall$$ x ϵ Df for least positive value T.

Example: If f: R→R is an odd function such that

(i) f(1+x) = 1+ f(x)

(ii) x² f(1/x) = f(x), x ≠ 0, Then find f(x).

Solution: Given that f(1 + x) = 1 + f(x)

x² f(1/x) = f(x), x ≠ 0

Replace x by x + 1, then

$${{(1+x)}^{2}}f\left( \frac{1}{1+x} \right)=f(x+1)$$,

Given condition f(1+x) = 1+ f(x)

$$f\left( \frac{1}{1+x} \right)=\frac{1+f(x)}{{{(1+x)}^{2}}}$$ … (1)

$$f\left( \frac{1}{1+x} \right)=f\left( 1-\frac{x}{1+x} \right)$$,

$$=1+f\left( -\frac{x}{1+x} \right)$$,

$$=1-f\left( \frac{x}{1+x} \right)$$   (∵ f(x) is odd)

$$=1-{{\left( \frac{x}{1+x} \right)}^{2}}f\left( \frac{1+x}{x} \right)$$,

$$=1-{{\left( \frac{x}{1+x} \right)}^{2}}f\left( 1+\frac{1}{x} \right)$$ … (2)

From (1) and (2) equations

$$\frac{1+f(x)}{{{(1+x)}^{2}}}=1-{{\left( \frac{x}{1+x} \right)}^{2}}\left( 1+\frac{f(x)}{{{x}^{2}}} \right)$$,

1 + f (x) = (1 + x)² – x² – f (x)

f(x) = x.