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Question Number 98842 by  M±th+et+s last updated on 16/Jun/20

$$letf\left(x\right)beadolvnomialofdegree4$$$$suchthatf\left(1\right)=1,f\left(2\right)=2,f\left(3\right)=3,f\left(4\right)=4$$$$thenf\left(6\right)=?$$

Commented by mr W last updated on 16/Jun/20

$$seealsoQ98268$$

Answered by maths mind last updated on 16/Jun/20

$$letp\left(x\right)=f\left(x\right)-x\Rightarrow P\left(x\right)\in {\mathbb{R}}_4\left[X\right]$$$$\Rightarrow p\left(1\right)=p\left(2\right)=p\left(3\right)=p\left(4\right)=0$$$$1,2,3,4arerootsofp$$$$p\left(x\right)=a(x-1)(x-2)(x-3)(x-4)$$$$f\left(x\right)=a(x-1)(x-2)(x-3)(x-4)+x$$$$f\left(6\right)=6+120a$$

Commented by  M±th+et+s last updated on 16/Jun/20

$$thankyousir$$

Answered by Rio Michael last updated on 17/Jun/20

$$\mathrm{Following}\mathrm{the}\mathrm{above}\mathrm{procedure},$$$$\mathrm{let}p\left(x\right)=f\left(x\right)-x$$$$\Rightarrow p\left(1\right)=f\left(1\right)-1=0$$$$p\left(2\right)=f\left(2\right)-2=0$$$$p\left(3\right)=f\left(3\right)-3=0$$$$p\left(4\right)=f\left(4\right)-4=0$$$$\Rightarrow f\left(x\right)=p\left(x\right)+x\mathrm{where}p\left(x\right)=(x-1)(x-2)(x-3)(x-4)$$$$\Rightarrow f\left(x\right)=(x-1)(x-2)(x-3)(x-4)+x$$$$f\left(6\right)=\left(5\right)\left(4\right)\left(3\right)\left(2\right)+6=126$$

Commented by mr W last updated on 17/Jun/20

$$4conditionsarenotenoughto$$$$uinquelydefinea4thdegree$$$$polynomial,thereforethereisno$$$$uniquesolutionforf\left(x\right).wecanonly$$$$say$$$$f\left(x\right)=k(x-1)(x-2)(x-3)(x-4)+x$$$$withk=constant\ne 0$$$$f\left(6\right)=120k+6$$

Commented by Rio Michael last updated on 17/Jun/20

$$\mathrm{thanks}\mathrm{sir},\mathrm{i}\mathrm{got}\mathrm{it}\mathrm{now}...$$$$$$

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