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Question Number 155281 by mnjuly1970 last updated on 28/Sep/21

$$$$$$f:[0,6]\to [-4,4]$$$$f\left(0\right)=0$$$$f\left(6\right)=4$$$$x,y⩾0,x+y⩽6$$$$f(x+y)=\frac{1}{4}\{f\left(x\right)\sqrt{16-{\left(f\left(y\right)\right)}^2}+f\left(y\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\}$$$$\therefore {(f\left(1\right)+f\left(3\right))}^2=?$$

Answered by Rasheed.Sindhi last updated on 29/Sep/21

$$f(x+y)=\frac{1}{4}\{f\left(x\right)\sqrt{16-{\left(f\left(y\right)\right)}^2}+f\left(y\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\}$$$$y=x:$$$$f\left(2x\right)=\frac{1}{4}\{f\left(x\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}+f\left(x\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\}$$$$f\left(2x\right)=\frac{1}{4}\left\{2f\left(x\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\right\}$$$$f\left(2x\right)=\frac{1}{2}\left\{f\left(x\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\right\}$$$$f\left(x\right)=\frac{1}{2}\left\{f\left(\frac{x}{2}\right)\sqrt{16-{\left(f\left(\frac{x}{2}\right)\right)}^2}\right\}$$$$f\left(6\right)=\frac{1}{2}\left\{f\left(3\right)\sqrt{16-{\left(f\left(3\right)\right)}^2}\right\}=4$$$$f\left(3\right)=t:t\sqrt{16-t^2}=8$$$$t^2(16-t^2)=64$$$$t^4-16t^2+64=0$$$${(t^2-8)}^2=0$$$$t=\pm 2\sqrt{2}$$$$f\left(3\right)=\pm 2\sqrt{2}$$$$f(x+y)=\frac{1}{4}\{f\left(x\right)\sqrt{16-{\left(f\left(y\right)\right)}^2}+f\left(y\right)\sqrt{16-{\left(f\left(x\right)\right)}^2}\}$$$$f\left(1\right)=?$$$${\mathrm{Can}}^{\prime }\mathrm{t}\mathrm{continue}.\mathrm{Pl}\mathrm{help}!$$

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