Question Number 215052 by Abdullahrussell last updated on 27/Dec/24
Answered by mr W last updated on 27/Dec/24
![f(1)=1 ⇒f(x)=(x−1)g(x)+1 f(2)=(2−1)g(2)+1=4 ⇒g(2)=3 ⇒g(x)=(x−2)h(x)+3 ⇒f(x)=(x−1)[(x−2)h(x)+3]+1 f(3)=(3−1)[(3−2)h(3)+3]+1=3 ⇒h(3)=−2 ⇒h(x)=(x−3)k(x)−2 ⇒f(x)=(x−1){(x−2)[(x−3)k(x)−2]+3}+1 say k(x)=ax+b f(x)=(x^3 −6x^2 +11x−6)(ax+b)−2x^2 +9x−6 f′(x)=a(x^3 −6x^2 +11x−6)+(3x^2 −12x+11)(ax+b)−4x+9 f′(1)=1 ⇒a+b=−2 f′(3)=1 ⇒3a+b=2 ⇒a=2, b=−4 f(x)=(x−1){2(x−2)[(x−3)(x−2)−1]+3}+1 or f(x)=2x^4 −16x^3 +44x^2 −47x+18 f(4)=22 ✓](https://www.tinkutara.com/question/Q215062.png)
$${f}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right){g}\left({x}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right){g}\left(\mathrm{2}\right)+\mathrm{1}=\mathrm{4}\:\Rightarrow{g}\left(\mathrm{2}\right)=\mathrm{3}\:\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left[\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3}\right]+\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}−\mathrm{1}\right)\left[\left(\mathrm{3}−\mathrm{2}\right){h}\left(\mathrm{3}\right)+\mathrm{3}\right]+\mathrm{1}=\mathrm{3}\:\Rightarrow{h}\left(\mathrm{3}\right)=−\mathrm{2}\:\Rightarrow{h}\left({x}\right)=\left({x}−\mathrm{3}\right){k}\left({x}\right)−\mathrm{2} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left\{\left({x}−\mathrm{2}\right)\left[\left({x}−\mathrm{3}\right){k}\left({x}\right)−\mathrm{2}\right]+\mathrm{3}\right\}+\mathrm{1} \\ $$$${say}\:{k}\left({x}\right)={ax}+{b} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}\right)\left({ax}+{b}\right)−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{6} \\ $$$${f}'\left({x}\right)={a}\left({x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{11}\right)\left({ax}+{b}\right)−\mathrm{4}{x}+\mathrm{9} \\ $$$${f}'\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{a}+{b}=−\mathrm{2} \\ $$$${f}'\left(\mathrm{3}\right)=\mathrm{1}\:\Rightarrow\mathrm{3}{a}+{b}=\mathrm{2} \\ $$$$\Rightarrow{a}=\mathrm{2},\:{b}=−\mathrm{4} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left\{\mathrm{2}\left({x}−\mathrm{2}\right)\left[\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)−\mathrm{1}\right]+\mathrm{3}\right\}+\mathrm{1} \\ $$$${or} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{4}} −\mathrm{16}{x}^{\mathrm{3}} +\mathrm{44}{x}^{\mathrm{2}} −\mathrm{47}{x}+\mathrm{18} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{22}\:\checkmark \\ $$
Commented by mr W last updated on 27/Dec/24
Answered by mr W last updated on 27/Dec/24
![alternative: say g(x)=f(x)−x≥0 f(1)−1=g(1)=0 f(3)−3=g(3)=0 f(2)−2=g(2)=4−2=2 ⇒g(x)=(x−1)(x−3)h(x) g(2)=(2−1)(2−3)h(2)=−h(2)=2 ⇒h(x)=(x−2)k(x)−2 g(x)=(x−1)(x−3)[(x−2)k(x)−2] say k(x)=ax+b g′(1)=0, g′(3)=0 ......](https://www.tinkutara.com/question/Q215064.png)
$${alternative}: \\ $$$${say}\:{g}\left({x}\right)={f}\left({x}\right)−{x}\geqslant\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)−\mathrm{1}={g}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)−\mathrm{3}={g}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)−\mathrm{2}={g}\left(\mathrm{2}\right)=\mathrm{4}−\mathrm{2}=\mathrm{2} \\ $$$$\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right){h}\left({x}\right) \\ $$$${g}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\right){h}\left(\mathrm{2}\right)=−{h}\left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\Rightarrow{h}\left({x}\right)=\left({x}−\mathrm{2}\right){k}\left({x}\right)−\mathrm{2} \\ $$$${g}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left[\left({x}−\mathrm{2}\right){k}\left({x}\right)−\mathrm{2}\right] \\ $$$${say}\:{k}\left({x}\right)={ax}+{b} \\ $$$${g}'\left(\mathrm{1}\right)=\mathrm{0},\:{g}'\left(\mathrm{3}\right)=\mathrm{0}\: \\ $$$$…… \\ $$