Consider a function f: R to R such that f(x + | vedclass

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(D) Given $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$.For the square root to be defined,we must have $f(x) - f^2(x) \geq 0$,which implies $f(x)(1

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periodic function

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(D) Given $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$.For the square root to be defined,we must have $f(x) - f^2(x) \geq 0$,which implies $f(x)(1 - f(x)) \geq 0$,so $0 \leq f(x) \leq 1$.Also,$f(x + a) \geq \frac{1}{2}$.Let $g(x) = f(x) - \frac{1}{2}$. Then $g(x + a) = \sqrt{\frac{1}{4} - g^2(x)}$.Squaring both sides,$g^2(x + a) = \frac{1}{4} - g^2(x)$,which implies $g^2(x + a) + g^2(x) = \frac{1}{4}$.Replacing $x$ with $x + a$,we get $g^2(x + 2a) + g^2(x + a) = \frac{1}{4}$.Comparing the two equations,$g^2(x + 2a) = g^2(x)$.Since $g(x) \geq 0$ (as $f(x) \geq \frac{1}{2}$),we have $g(x + 2a) = g(x)$.Substituting back,$f(x + 2a) - \frac{1}{2} = f(x) - \frac{1}{2}$,so $f(x + 2a) = f(x)$.Thus,$f(x)$ is a periodic function with period $2a$.

Consider a function $f: R \to R$ such that $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$,where $a$ is a real constant. Then $f(x)$ must be

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