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Question Number 174806 by cortano1 last updated on 11/Aug/22

$$Givenf\left(x\right)=x^4+{ax}^3+{bx}^2+cx+d$$$$wherea,b,canddarerealnumber$$$$supposethegraphf\left(x\right)intersects$$$$thegraphofy=2x-1atx=1,2,3.$$$$Findthevalueoff\left(0\right)+f\left(4\right).$$

Commented by Rasheed.Sindhi last updated on 11/Aug/22

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by floor(10²Eta[1]) last updated on 11/Aug/22

$$\Rightarrow 1+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=1\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\left(\mathrm{I}\right)$$$$16+8\mathrm{a}+4\mathrm{b}+2\mathrm{c}+\mathrm{d}=3\Rightarrow 8\mathrm{a}+4\mathrm{b}+2\mathrm{c}+\mathrm{d}=-13\left(\mathrm{II}\right)$$$$81+27\mathrm{a}+9\mathrm{b}+3\mathrm{c}+\mathrm{d}=5\left(\mathrm{III}\right)$$$$$$$$\Rightarrow \left(\mathrm{II}\right)-\left(\mathrm{I}\right)=7\mathrm{a}+3\mathrm{b}+\mathrm{c}=-13\left(\mathrm{IV}\right)$$$$\left(\mathrm{III}\right)-\left(\mathrm{I}\right)=26\mathrm{a}+8\mathrm{b}+2\mathrm{c}=-76\left(\mathrm{V}\right)$$$$\left(\mathrm{V}\right)-2\left(\mathrm{IV}\right)=12\mathrm{a}+2\mathrm{b}=-50\Rightarrow 6\mathrm{a}+\mathrm{b}=-25\left(\mathrm{VI}\right)$$$$$$$$\mathrm{f}\left(0\right)+\mathrm{f}\left(4\right)=256+64\mathrm{a}+16\mathrm{b}+4\mathrm{c}+2\mathrm{d}$$$$=256+48\mathrm{a}+8\mathrm{b}+(16\mathrm{a}+8\mathrm{b}+4\mathrm{c}+2\mathrm{d})\to 2\left(\mathrm{II}\right)$$$$=256+48\mathrm{a}+8\mathrm{b}-26=230+8(6\mathrm{a}+\mathrm{b})\to \left(\mathrm{VI}\right)$$$$=230-8\left(25\right)=30$$

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