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Question Number 191151 by TUN last updated on 19/Apr/23

$$f\left(x\right)+f(1-x)=x^2$$$$=>f\left(x\right)=¿$$

Answered by Rasheed.Sindhi last updated on 19/Apr/23

$$f\left(x\right)+f(1-x)=x^2.......\left(i\right)$$$$Replacingxby1-x:$$$$f(1-x)+f(1-(1-x))={(1-x)}^2$$$$f(1-x)+f\left(x\right)={(1-x)}^2......\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right):x^2={(1-x)}^2$$$$x^2=1-2x+x^2$$$$1-2x=0\Rightarrow x=\frac{1}{2}$$$$xisconstant$$$$f\left(\frac{1}{2}\right)+f(1-\frac{1}{2})={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$$$$2f\left(\frac{1}{2}\right)=\frac{1}{4}\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{8}$$

Commented by mr W last updated on 19/Apr/23

$$butinf\left(x\right)+f(1-x)=x^{2}xisnota$$$$constantbutavariable(\in R).$$$$soithinktheansweristhat$$$$f\left(x\right)satisfyingf\left(x\right)+f(1-x)=x^2$$$${doesn}^{\prime }texist.$$

Commented by Rasheed.Sindhi last updated on 19/Apr/23

$$\mathrm{Very}\mathrm{right}\mathrm{sir}!\mathbf{Thanks}!$$

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