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Question Number 152671 by liberty last updated on 31/Aug/21

Commented by MJS_new last updated on 31/Aug/21

$$f\left(1\right)=5$$$$f\left(2\right)=f(1+1)=f\left(1\right)+6\times 1+4=15$$$$f\left(3\right)=f(2+1)=f\left(2\right)+6\times 2+4=31$$$$\mathrm{but}$$$$f\left(3\right)=f\left(3\times 1\right)=4f\left(1\right)+1=21$$$${\mathrm{something}}^{\prime }\mathrm{s}\mathrm{wrong}$$

Commented by liberty last updated on 31/Aug/21

$$\mathrm{from}\mathrm{f}(\mathrm{x}+1)=\mathrm{f}\left(\mathrm{x}\right)+6\mathrm{x}+4$$$$\mathrm{x}=1\Rightarrow \mathrm{f}\left(2\right)=5+6+4=15$$$$\mathrm{x}=2\Rightarrow \mathrm{f}\left(3\right)=15+12+4=31$$$$\mathrm{x}=3\Rightarrow \mathrm{f}\left(4\right)=31+18+4=53$$$$\mathrm{x}=4\Rightarrow \mathrm{f}\left(5\right)=53+24+4=91$$$$\mathrm{x}=5\Rightarrow \mathrm{f}\left(6\right)=91+30+4=125$$$$\mathrm{therefore}\mathrm{fix}\mathrm{the}\mathrm{question}$$$$\mathrm{something}\mathrm{wrong}$$

Answered by Rasheed.Sindhi last updated on 31/Aug/21

$$f\left(1\right)=5;\underset{\left(1\right)}{\underbrace{f\left(3x\right)=4f\left(x\right)+1}};\underset{\left(2\right)}{\underbrace{f(x+1)=f\left(x\right)+6x+4}};f\left(x\right)=?$$$${}^{\left(2\right)\Rightarrow }f(1+1)=f\left(1\right)+6\left(1\right)+4=5+6+4=15$$$$f\left(2\right)=15$$$${}^{\left(1\right)\Rightarrow }f\left(3\times 2\right)=4f\left(2\right)+1=4\left(15\right)+1=61$$$$f\left(6\right)=61$$

Answered by chengulapetrom last updated on 31/Aug/21

$$f\left(3\left(1\right)\right)=4f\left(1\right)+1$$$$f\left(3\right)=21$$$$f(1+1)=f\left(1\right)+6+4$$$$f\left(2\right)=15$$$$f\left(3\times 2\right)=4f\left(2\right)+1$$$$f\left(6\right)=4\times 15+1$$$$f\left(6\right)=61$$

Answered by Olaf_Thorendsen last updated on 31/Aug/21

$$f\left(6\right)=f\left(3\times 2\right)=4f\left(2\right)+1$$$$=4f(1+1)+1=4(f\left(1\right)+6\times 1+4)+1$$$$=4(5+10)+1=61$$

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