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Question Number 200167 by hardmath last updated on 15/Nov/23

$$\mathrm{Given}\mathrm{f}:\mathbb{R}\to \mathbb{R}\mathrm{is}\mathrm{a}\mathrm{quadratic}\mathrm{polynomial}$$$$\mathrm{f}\left(1\right)=1,\mathrm{f}\left(2\right)=\frac{1}{2}\mathrm{and}\mathrm{f}\left(3\right)=\frac{1}{3}$$$$\mathrm{Find}:\mathrm{f}\left(4\right)=?$$

Answered by witcher3 last updated on 15/Nov/23

$$\iff \mathrm{f}\left(1\right)-1=2\mathrm{f}\left(2\right)-1=3\mathrm{f}\left(3\right)-1=0$$$$\mathrm{let}\mathrm{xf}\left(\mathrm{x}\right)-1=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow \mathrm{deg}\left(\mathrm{g}\right)=3$$$$\mathrm{g}\left(1\right)=\mathrm{g}\left(2\right)=\mathrm{g}\left(3\right)=0$$$$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\mathrm{a}(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)$$$$\mathrm{xf}\left(\mathrm{x}\right)=1+\mathrm{g}\left(\mathrm{x}\right)$$$$\Rightarrow 1+\mathrm{g}\left(0\right)=0\Rightarrow -6\mathrm{a}+1=0$$$$\mathrm{a}=\frac{1}{6}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}(\frac{1}{6}(\mathrm{x}-3)(\mathrm{x}-2)(\mathrm{x}-1)+1)$$$$={\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{6}-\mathrm{x}+\frac{11}{6}=\mathrm{f}\left(\mathrm{x}\right)$$$$$$

Answered by Rasheed.Sindhi last updated on 15/Nov/23

$$letf\left(x\right)={ax}^2+bx+c$$$$f\left(1\right)=a+b+c=1$$$$f\left(2\right)=4a+2b+c=\frac{1}{2}$$$$f\left(3\right)=9a+3b+c=\frac{1}{3}$$$$f\left(2\right)-f\left(1\right)=3a+b=-\frac{1}{2}$$$$3(f\left(2\right)-f\left(1\right))=9a+3b=-\frac{3}{2}$$$$f\left(3\right)-3(f\left(2\right)-f\left(1\right))=c=\frac{1}{3}-(-\frac{3}{2})=\frac{11}{6}$$$$f\left(1\right)\Rightarrow a+b=1-\frac{11}{6}=-\frac{5}{6}$$$$2a+2b=-\frac{5}{3}..................\left(i\right)$$$$f\left(2\right)\Rightarrow 4a+2b=\frac{1}{2}-\frac{11}{6}=-\frac{4}{3}......\left(ii\right)$$$$\left(ii\right)-\left(ii\right):2a=-\frac{4}{3}-(-\frac{5}{3})=\frac{1}{3}$$$$a=\frac{1}{6}$$$$f\left(1\right):\frac{1}{6}+b+\frac{11}{6}=1\Rightarrow b=1-2=-1$$$$f\left(4\right)=16a+4b+c=16\left(\frac{1}{6}\right)+4(-1)+\frac{11}{6}$$$$=\frac{16-24+11}{6}=\frac{3}{6}=\frac{1}{2}✓$$

Answered by Rasheed.Sindhi last updated on 15/Nov/23

$$letf\left(x\right)={ax}^2+bx+c$$$$\overline{)\begin{array}{cccc}1)& a& b& c\\ & & a& a+b\\ & a& a+b& f\left(1\right)=a+b+c=1\end{array}}$$$$a=1-b-c$$$$\overline{)\begin{array}{cccc}2)& 1-b-c& b& c\\ & & 2-2b-2c& 4-2b-4c\\ & 1-b-c& 2-b-2c& f\left(2\right)=4-2b-3c=\frac{1}{2}\end{array}}$$$$-4+2b+3c=-\frac{1}{2}\Rightarrow b=(\frac{7}{2}-3c)/2=\frac{7}{4}-\frac{3}{2}c$$$$a=1-b-c=1-(\frac{7}{4}-\frac{3}{2}c)-c=-\frac{3}{4}+\frac{\mathrm{c}}{2}$$$$\overline{)\begin{array}{cccc}3)& -\frac{3}{4}+\frac{c}{2}& \frac{7}{4}-\frac{3}{2}c& c\\ & & -\frac{9}{4}+\frac{3c}{2}& -\frac{3}{2}\\ & -\frac{3}{4}+\frac{c}{2}& -\frac{1}{2}& f\left(3\right)=-\frac{3}{2}+c=\frac{1}{3}\end{array}}$$$$$$$$c=\frac{1}{3}+\frac{3}{2}=\frac{11}{6}$$$$a=-\frac{3}{4}+\frac{c}{2}=-\frac{3}{4}+\frac{1}{2}\cdot \frac{11}{6}=\frac{-9+11}{12}=\frac{2}{12}=\frac{1}{6}$$$$b=\frac{7}{4}-\frac{3}{2}c=\frac{7}{4}-\frac{3}{2}\left(\frac{11}{6}\right)=\frac{7-11}{4}=-1$$$$$$$$f\left(x\right)=\frac{1}{6}{x}^2-x+\frac{11}{6}$$$$f\left(4\right)=?$$$$\overline{)\begin{array}{cccc}4)& \frac{1}{6}& -1& \frac{11}{6}\\ & & \frac{2}{3}& -\frac{4}{3}\\ & \frac{1}{6}& -\frac{1}{3}& f\left(4\right)=\frac{1}{2}\end{array}}$$$$$$

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