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Question Number 132031 by mathlove last updated on 10/Feb/21

Commented by kaivan.ahmadi last updated on 10/Feb/21

$$p\left(x\right)=f\left(x\right)(x-1)+3=g\left(x\right)(x+1)+7$$$$p\left(1\right)=3,p(-1)=7$$$$p\left(x\right)=t\left(x\right)(x^2-1)+k\left(x\right)$$$$p\left(1\right)=k\left(1\right)\Rightarrow k\left(1\right)=3\Rightarrow (1,3)\in k$$$$p(-1)=k(-1)=7\Rightarrow (-1,7)\in k$$$$k\left(x\right)-3=\frac{7-3}{-1-1}(x-1)\Rightarrow k\left(x\right)-3=-2(x-1)$$$$\Rightarrow k\left(x\right)=-2x+5$$$$$$

Commented by kaivan.ahmadi last updated on 10/Feb/21

$$k\left(1\right)=3,k(-1)=7$$$$k\left(x\right)=ax+b$$$$\{\begin{array}{l}x=1\Rightarrow a+b=3\\ x=-1\Rightarrow -a+b=7\end{array}\Rightarrow b=5,a=-2$$$$\Rightarrow k\left(x\right)=-2x+5$$

Commented by mathlove last updated on 11/Feb/21

$$thankssir$$

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