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Question Number 193962 by Abdullahrussell last updated on 24/Jun/23

Answered by deleteduser1 last updated on 24/Jun/23

⇒f(x)=Q(x)(x−a)(x−b)+R⇒f(a)=R,f(b)=R ⇒((f(a)(x−b)−f(b)(x−a))/(a−b))=((R(x−b−x+a))/(a−b))=R

$$\Rightarrow{f}\left({x}\right)={Q}\left({x}\right)\left({x}−{a}\right)\left({x}−{b}\right)+{R}\Rightarrow{f}\left({a}\right)={R},{f}\left({b}\right)={R} \\ $$$$\Rightarrow\frac{{f}\left({a}\right)\left({x}−{b}\right)−{f}\left({b}\right)\left({x}−{a}\right)}{{a}−{b}}=\frac{{R}\left({x}−{b}−{x}+{a}\right)}{{a}−{b}}={R} \\ $$

Answered by cortano12 last updated on 24/Jun/23

x=a⇒f(a)= ma+n x=b⇒f(b)=mb+n ⇒ { ((m=((f(a)−f(b))/(a−b)))),((n=f(a)−a(((f(a)−f(b))/(a−b))))) :} ⇒n=((af(a)−bf(a)−af(a)+af(b))/(a−b)) ⇒n= ((af(b)−bf(a))/(a−b)) (((f(a)−f(b))/(a−b)))x+((af(b)−bf(a))/(a−b)) determinant (((R=((f(a)(x−b)−f(b)(x−a))/(a−b)))))

$$\:\:\: \\ $$$$ {x}={a}\Rightarrow\mathrm{f}\left({a}\right)=\:{ma}+{n} \\ $$$$\:{x}={b}\Rightarrow\mathrm{f}\left({b}\right)={mb}+{n}\: \\ $$$$\:\:\Rightarrow\begin{cases}{{m}=\frac{{f}\left({a}\right)−{f}\left({b}\right)}{{a}−{b}}}\\{{n}={f}\left({a}\right)−{a}\left(\frac{{f}\left({a}\right)−{f}\left({b}\right)}{{a}−{b}}\right)}\end{cases} \\ $$$$\:\:\Rightarrow{n}=\frac{{af}\left({a}\right)−{bf}\left({a}\right)−{af}\left({a}\right)+{af}\left({b}\right)}{{a}−{b}} \\ $$$$\:\:\Rightarrow{n}=\:\frac{{af}\left({b}\right)−{bf}\left({a}\right)}{{a}−{b}} \\ $$$$\: \\ $$$$ \left(\frac{{f}\left({a}\right)−{f}\left({b}\right)}{{a}−{b}}\right){x}+\frac{{af}\left({b}\right)−{bf}\left({a}\right)}{{a}−{b}} \\ $$$$\:\:\begin{array}{|c|}{\mathrm{R}=\frac{{f}\left({a}\right)\left({x}−{b}\right)−{f}\left({b}\right)\left({x}−{a}\right)}{{a}−{b}}}\\\hline\end{array} \\ $$$$ \\ $$$$ \\ $$

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