Find-a-polynomial-f-x-which-satisfy-the-following-conditions-i-f-x-of-degree-4-ii-x-1-is-a-factor-of-f-x-and-f-x-iii-f-0-3-and-f-0-5-iv-when-f-x-is-divided-by-x-2-the-remain

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Question Number 105707 by ZiYangLee last updated on 31/Jul/20

$$\mathrm{Find}\mathrm{a}\mathrm{polynomial}\mathrm{f}\left(\mathrm{x}\right)\mathrm{which}\mathrm{satisfy}$$$$\mathrm{the}\mathrm{following}\mathrm{conditions}:$$$$\mathrm{i})\mathrm{f}\left(\mathrm{x}\right)\mathrm{of}\mathrm{degree}4$$$$\mathrm{ii})(\mathrm{x}-1)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)\mathrm{and}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)$$$$\mathrm{iii})\mathrm{f}\left(0\right)=3\mathrm{and}{\mathrm{f}}^{\prime }\left(0\right)=-5$$$$\mathrm{iv})\mathrm{when}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{divided}\mathrm{by}(\mathrm{x}-2),\mathrm{the}$$$$\mathrm{remainder}\mathrm{is}13.$$

Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

$$\mathrm{let}$$$$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}-1)}^2({\mathrm{ax}}^2+\mathrm{bx}+3)$$$$={\mathbf{ax}}^4+{\mathbf{x}}^3(\mathbf{b}-2\mathbf{a})+{\mathbf{x}}^2(\mathbf{a}-2\mathbf{b}+3)+\mathbf{x}(\mathbf{b}-6)+3$$$${\mathbf{f}}^{\prime }\left(\mathbf{x}\right)=4{\mathbf{ax}}^3+3{\mathbf{x}}^2(\mathbf{b}-2\mathbf{a})+2\mathbf{x}(\mathbf{a}-2\mathbf{b}+3)+(\mathbf{b}-6)$$$${\mathbf{f}}^{{}^{\prime }}\left(0\right)=-5\Rightarrow \mathbf{b}-6=-5\Rightarrow \mathbf{b}=1$$$$\mathbf{f}\left(2\right)=13\Rightarrow 2\mathbf{a}+\mathbf{b}=5\Rightarrow \mathbf{a}=2$$$$\therefore \mathbf{f}\left(\mathbf{x}\right)=2{\mathbf{x}}^4-3{\mathbf{x}}^3+3{\mathbf{x}}^2-5\mathbf{x}+3$$

Answered by bobhans last updated on 31/Jul/20

$$f\left(x\right)=(x-1)(2x^3-x^2+2x-3)\therefore$$$$letf\left(x\right)=(x-1)({ax}^3+{bx}^2+cx+d)$$$$\left(1\right)f\left(0\right)=-d=3;d=-3$$$$\left(2\right)f{}^{\prime }\left(x\right)=(x-1)(3{ax}^2+2bx+c)+({ax}^3+{bx}^2+cx-3)$$$$f{}^{\prime }\left(0\right)=-c-3=-5;c=2$$$$\left(3\right)f{}^{\prime }\left(1\right)=0\to a+b+2-3=0;b=1-a$$$$\left(4\right)f\left(x\right)=(x-1)({ax}^3+(1-a)x^2+2x-3)$$$$f\left(2\right)=8a+4-4a+1=13;4a=8;a=2$$$$thenb=-1$$

Commented by bemath last updated on 31/Jul/20

$$cooll\dots .$$

Answered by floor(10²Eta[1]) last updated on 31/Jul/20

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={4\mathrm{ax}}^3+{3\mathrm{bx}}^2+2\mathrm{cx}+\mathrm{d}$$$$\mathrm{by}\mathrm{info}2:$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{A}\left(\mathrm{x}\right)(\mathrm{x}-1)$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{A}}^{\prime }\left(\mathrm{x}\right)(\mathrm{x}-1)+\mathrm{A}\left(\mathrm{x}\right),\mathrm{but}\mathrm{since}$$$$\mathrm{x}-1\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right),\mathrm{so}\mathrm{x}-1\mathrm{has}\mathrm{to}$$$$\mathrm{be}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{A}\left(\mathrm{x}\right)\Rightarrow \mathrm{A}\left(\mathrm{x}\right)=\mathrm{B}\left(\mathrm{x}\right)(\mathrm{x}-1)$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{B}\left(\mathrm{x}\right){(\mathrm{x}-1)}^2$$$$\mathrm{by}\mathrm{info}3\mathrm{and}4:$$$$\mathrm{f}\left(0\right)=\mathrm{e}=3,{\mathrm{f}}^{\prime }\left(0\right)=\mathrm{d}=-5,\mathrm{f}\left(2\right)=13$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2-5\mathrm{x}+3$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={4\mathrm{ax}}^3+{3\mathrm{bx}}^2+2\mathrm{cx}-5$$$$\Rightarrow \mathrm{f}\left(1\right)=\mathrm{a}+\mathrm{b}+\mathrm{c}-5+3=0$$$${\mathrm{f}}^{\prime }\left(1\right)=4\mathrm{a}+3\mathrm{b}+2\mathrm{c}-5=0$$$$\mathrm{and}\mathrm{f}\left(2\right)=16\mathrm{a}+8\mathrm{b}+4\mathrm{c}-10+3=13$$$$\left(\mathrm{I}\right):\mathrm{a}+\mathrm{b}+\mathrm{c}=2$$$$\left(\mathrm{II}\right):4\mathrm{a}+3\mathrm{b}+2\mathrm{c}=5$$$$\left(\mathrm{III}\right):16\mathrm{a}+8\mathrm{b}+4\mathrm{c}=20\Rightarrow 4\mathrm{a}+2\mathrm{b}+\mathrm{c}=5$$$$\mathrm{solving}\mathrm{this}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{we}\mathrm{get}:$$$$\mathrm{a}=2,\mathrm{b}=-3,\mathrm{c}=3$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={2\mathrm{x}}^4-{3\mathrm{x}}^3+{3\mathrm{x}}^2-5\mathrm{x}+3.$$$$$$

Commented by bemath last updated on 31/Jul/20

$$f\left(2\right)=16-8+8-10+3=9$$$$wrongsir$$

Commented by floor(10²Eta[1]) last updated on 31/Jul/20

$$\mathrm{oh}\mathrm{i}\mathrm{actually}\mathrm{missed}\mathrm{the}\mathrm{information}4$$

Answered by 1549442205PVT last updated on 31/Jul/20

$$\mathrm{General}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{a}\mathrm{polynomyal}\mathrm{of}$$$$\mathrm{degree}4\mathrm{is}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}.\mathrm{Then}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)={4\mathrm{ax}}^3+{3\mathrm{bx}}^2+2\mathrm{cx}+\mathrm{d},$$$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{iii})\mathrm{we}\mathrm{get}:$$$$\mathrm{f}\left(0\right)=3\iff \mathrm{e}=3,\mathrm{f}{}^{\prime }\left(0\right)=-5\iff \mathrm{d}=-5.\mathrm{Hence}$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2-5\mathrm{x}+3$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)={4\mathrm{ax}}^3+{3\mathrm{bx}}^2+2\mathrm{cx}-5$$$$\mathrm{From}\mathrm{ii})\mathrm{we}\mathrm{get}\mathrm{f}\left(1\right)=0\iff \mathrm{a}+\mathrm{b}+\mathrm{c}-2=0\left(1\right)$$$$\mathrm{f}{}^{\prime }\left(1\right)=0\iff 4\mathrm{a}+3\mathrm{b}+2\mathrm{c}-5=0\left(2\right)$$$$\mathrm{From}\mathrm{iv})\mathrm{we}\mathrm{get}\mathrm{f}\left(2\right)=13\iff 16\mathrm{a}+8\mathrm{b}+4\mathrm{c}-7=13\left(3\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{get}$$$$\{\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}-2=0\\ 4\mathrm{a}+3\mathrm{b}+2\mathrm{c}-5=0\\ 16\mathrm{a}+8\mathrm{b}+4\mathrm{c}-7=13\end{array}\iff \{\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}=2\left(4\right)\\ 4\mathrm{a}+3\mathrm{b}+2\mathrm{c}=5\\ 4\mathrm{a}+2\mathrm{b}+\mathrm{c}=5\left(6\right)\end{array}\left(5\right)$$$$\mathrm{Substract}\left(5\right)\mathrm{and}\left(6\right)\mathrm{we}\mathrm{getb}+\mathrm{c}=0\left(7\right),$$$$\mathrm{replace}\mathrm{in}\left(4\right)\mathrm{we}\mathrm{obtain}\mathrm{a}=2,\mathrm{replace}$$$$\mathrm{into}\left(6\right)\mathrm{we}\mathrm{get}2\mathrm{b}+\mathrm{c}=-3\left(8\right).\mathrm{From}$$$$\left(7\right),\left(8\right)\mathrm{we}\mathrm{obtain}\mathrm{b}=-3,\mathrm{c}=3.$$$$\mathbf{Thus},\mathbf{f}\left(\mathbf{x}\right)=2{\mathbf{x}}^4-3{\mathbf{x}}^3+3{\mathbf{x}}^2-5\mathbf{x}+3$$

Answered by malwaan last updated on 31/Jul/20

$$f\left(x\right)={ax}^4+{bx}^3+{cx}^2+dx+e$$$$f\left(1\right)=a+b+c+d+e=0\dots 1$$$$f^{\prime }\left(1\right)=4a+3b+2c+d=0\dots 2$$$$f\left(0\right)=3\Rightarrow e=3\dots ..3$$$$f^{\prime }\left(0\right)=-5\Rightarrow d=-5\dots .4$$$$f\left(2\right)=16a+8b+4c+2d+e=13..5$$$$1\Rightarrow a+b+c-5+3=0$$$$\Rightarrow a+b+c=2\dots .6$$$$2\Rightarrow 4a+3b+2c-5=0$$$$\Rightarrow 4a+3b+2c=5\dots .7$$$$5\Rightarrow 16a+8b+4c-10+3=13$$$$\Rightarrow 16a+8b+4c=20$$$$\Rightarrow 4a+2b+c=5\dots 8$$$$solvethesystemofequations$$$$6;7;8weget$$$$c=-3;b=3;a=2$$$$thepolynomialis$$$$f\left(x\right)=2x^4-3x^3+3x^2-5x+3$$

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