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Question Number 174806 by cortano1 last updated on 11/Aug/22
Given f(x)=x^4 +ax^3 +bx^2 +cx+d where a,b,c and d are real number suppose the graph f(x) intersects the graph of y=2x−1 at x=1,2,3. Find the value of f(0)+f(4).
$${Given}\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${where}\:{a},{b},{c}\:{and}\:{d}\:{are}\:{real}\:{number} \\ $$$${suppose}\:{the}\:{graph}\:{f}\left({x}\right)\:{intersects} \\ $$$${the}\:{graph}\:{of}\:{y}=\mathrm{2}{x}−\mathrm{1}\:{at}\:{x}=\mathrm{1},\mathrm{2},\mathrm{3}. \\ $$$$\:{Find}\:{the}\:{value}\:{of}\:\:{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right). \\ $$
Commented by Rasheed.Sindhi last updated on 11/Aug/22
Q#158988
$${Q}#\mathrm{158988} \\ $$
Answered by floor(10²Eta[1]) last updated on 11/Aug/22
⇒1+a+b+c+d=1⇒a+b+c+d=0 (I) 16+8a+4b+2c+d=3⇒8a+4b+2c+d=−13 (II) 81+27a+9b+3c+d=5 (III) ⇒(II)−(I)=7a+3b+c=−13(IV) (III)−(I)=26a+8b+2c=−76 (V) (V)−2(IV)=12a+2b=−50⇒6a+b=−25 (VI) f(0)+f(4)=256+64a+16b+4c+2d =256+48a+8b+(16a+8b+4c+2d) →2(II) =256+48a+8b−26=230+8(6a+b) →(VI) =230−8(25)=30
$$\Rightarrow\mathrm{1}+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{1}\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{0}\:\left(\mathrm{I}\right) \\ $$$$\mathrm{16}+\mathrm{8a}+\mathrm{4b}+\mathrm{2c}+\mathrm{d}=\mathrm{3}\Rightarrow\mathrm{8a}+\mathrm{4b}+\mathrm{2c}+\mathrm{d}=−\mathrm{13}\:\left(\mathrm{II}\right) \\ $$$$\mathrm{81}+\mathrm{27a}+\mathrm{9b}+\mathrm{3c}+\mathrm{d}=\mathrm{5}\:\left(\mathrm{III}\right) \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{II}\right)−\left(\mathrm{I}\right)=\mathrm{7a}+\mathrm{3b}+\mathrm{c}=−\mathrm{13}\left(\mathrm{IV}\right) \\ $$$$\left(\mathrm{III}\right)−\left(\mathrm{I}\right)=\mathrm{26a}+\mathrm{8b}+\mathrm{2c}=−\mathrm{76}\:\left(\mathrm{V}\right) \\ $$$$\left(\mathrm{V}\right)−\mathrm{2}\left(\mathrm{IV}\right)=\mathrm{12a}+\mathrm{2b}=−\mathrm{50}\Rightarrow\mathrm{6a}+\mathrm{b}=−\mathrm{25}\:\left(\mathrm{VI}\right) \\ $$$$ \\ $$$$\mathrm{f}\left(\mathrm{0}\right)+\mathrm{f}\left(\mathrm{4}\right)=\mathrm{256}+\mathrm{64a}+\mathrm{16b}+\mathrm{4c}+\mathrm{2d} \\ $$$$=\mathrm{256}+\mathrm{48a}+\mathrm{8b}+\left(\mathrm{16a}+\mathrm{8b}+\mathrm{4c}+\mathrm{2d}\right)\:\rightarrow\mathrm{2}\left(\mathrm{II}\right) \\ $$$$=\mathrm{256}+\mathrm{48a}+\mathrm{8b}−\mathrm{26}=\mathrm{230}+\mathrm{8}\left(\mathrm{6a}+\mathrm{b}\right)\:\rightarrow\left(\mathrm{VI}\right) \\ $$$$=\mathrm{230}−\mathrm{8}\left(\mathrm{25}\right)=\mathrm{30} \\ $$

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