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Question Number 12889 by sin (x) last updated on 05/May/17

$$f(x-1)=\frac{2}{3}+f\left(x\right)$$$$f\left(0\right)=36-f\left(21\right)$$$$\Rightarrow f\left(0\right)=?$$

Commented by prakash jain last updated on 06/May/17

$$f\left(x\right)-f(x-1)=-\frac{2}{3}$$$$f\left(0\right)=36-f\left(21\right)....\left(1\right)$$$$f\left(1\right)-f\left(0\right)=-\frac{2}{3}$$$$..$$$$f\left(21\right)-f\left(20\right)=-\frac{2}{3}....\left(22\right)$$$$adding\left(1\right)to\left(22\right)$$$$f\left(21\right)=-\frac{2}{3}\times 21+36-f\left(21\right)$$$$2f\left(21\right)=22$$$$f\left(21\right)=11$$$$substitutingin\left(1\right)$$$$f\left(0\right)=36-11$$$$f\left(0\right)=25$$

Answered by prakash jain last updated on 06/May/17

$$f\left(0\right)=25$$

Answered by Joel577 last updated on 07/May/17

$$f\left(x\right)-f(x-1)=-\frac{2}{3}$$$$\mathrm{Let}f\left(x\right)=ax+b$$$$f(x-1)=a(x-1)+b=ax-a+b$$$$(ax+b)-ax+a-b=-\frac{2}{3}$$$$a=-\frac{2}{3}$$$$f\left(21\right)+f\left(0\right)=36$$$$(-\frac{2}{3}.21+b)+(-\frac{2}{3}.0+b)=36$$$$-14+2b=36$$$$b=\frac{36+14}{2}=25$$$$f\left(0\right)=-\frac{2}{3}.0+25=25$$

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