Menu Close

Given-t-x-ax-4-bx-2-x-5-where-a-and-b-are-constant-If-t-4-3-then-t-4-




Question Number 128176 by liberty last updated on 05/Jan/21
Given t(x)=ax^4 +bx^2 +x+5 ; where a and b  are constant. If t(−4)=3 then t(4)=?
$$\mathrm{Given}\:\mathrm{t}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{x}+\mathrm{5}\:;\:\mathrm{where}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b} \\ $$$$\mathrm{are}\:\mathrm{constant}.\:\mathrm{If}\:\mathrm{t}\left(−\mathrm{4}\right)=\mathrm{3}\:\mathrm{then}\:\mathrm{t}\left(\mathrm{4}\right)=? \\ $$
Answered by bemath last updated on 05/Jan/21
(⇒) t(x)=ax^4 +bx^2 +x+5   (⇒)t(4)=256a+16b+9  we have t(−4)=256a+16b+1  so t(−4) gives us 256a+16b=2  so we find t(4)=256a+16b+9=2+9=11
$$\left(\Rightarrow\right)\:\mathrm{t}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{x}+\mathrm{5}\: \\ $$$$\left(\Rightarrow\right)\mathrm{t}\left(\mathrm{4}\right)=\mathrm{256a}+\mathrm{16b}+\mathrm{9} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{t}\left(−\mathrm{4}\right)=\mathrm{256a}+\mathrm{16b}+\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{t}\left(−\mathrm{4}\right)\:\mathrm{gives}\:\mathrm{us}\:\mathrm{256a}+\mathrm{16b}=\mathrm{2} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{find}\:\mathrm{t}\left(\mathrm{4}\right)=\mathrm{256a}+\mathrm{16b}+\mathrm{9}=\mathrm{2}+\mathrm{9}=\mathrm{11} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *