Find all functions
$f : [0, \infty ) \to \mathbb{R},$ that is, an expression for $f(x),$ such that for all $x,y \in [0,\infty)$ we have $f(x+2y)+2f^2(y)+x = f^2(x+2y)+2f(y)-\frac{1}{4}.$
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Determine the value of
$f$for some simple cases. -
What about the value of
$f(0)?$ -
Why not try getting an equation of
$f$with a single variable? -
… by substituting a suitable value for
$y.$
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By setting
$x=y=0,$$$\begin{aligned} f(0) + 2f^2(0) &= f^2(0) + 2f(0) - \frac{1}{4} \\ f^2(0) - f(0) + \frac{1}{4} &= 0 \\ \big(f(0) - \frac{1}{2}\big)^2 &= 0, \\ \end{aligned}$$which gives us
$f(0) = \frac{1}{2}.$Now, we can obtain$f(x)$by setting$y=0:$$$\begin{aligned} f(x)+2f^2(0)+x &= f^2(x)+2f(0)-\frac{1}{4} \\ f^2(x)-f(x)+\frac{1}{4}-x &= 0.\\ \end{aligned}$$Solving this quadratic equation gives us
$f(x) = \frac{1}{2} \pm \sqrt{x}.$