f(x) = x² +3x + 2 dengan x real, maka f(2x+1) -f(2x) + f(x²+1) =... ?

[tex]f(x) = {x}^{2} + 3x + 2 \\ f(2x + 1) = ({2x + 1})^{2} + 3(2x + 1) + 2 \\ = 4 {x}^{2} + 4x + 1 + 6x + 3 + 2 \\ = 4 {x}^{2} + 10x + 6 \\ f(2x) = (2 {x})^{2} + 3(2x) + 2 \\ = 4 {x}^{2} + 6x + 2 \\ f( {x}^{2} + 1) = {( {x}^{2} + 1) }^{2} + 3( {x}^{2} + 1) + 2 \\ = {x}^{4} + 2 {x}^{2} + 1 + 3 {x}^{2} + 3 + 2 \\ = {x}^{4} + 5 {x}^{2} + 6[/tex]
[tex]f(x) = {x}^{2} + 3x + 2 \\ f(2x + 1) = ({2x + 1})^{2} + 3(2x + 1) + 2 \\ = 4 {x}^{2} + 4x + 1 + 6x + 3 + 2 \\ = 4 {x}^{2} + 10x + 6 \\ f(2x) = (2 {x})^{2} + 3(2x) + 2 \\ = 4 {x}^{2} + 6x + 2 \\ f( {x}^{2} + 1) = {( {x}^{2} + 1) }^{2} + 3( {x}^{2} + 1) + 2 \\ = {x}^{4} + 2 {x}^{2} + 1 + 3 {x}^{2} + 3 + 2 \\ = {x}^{4} + 5 {x}^{2} + 6 \\ \\ f(2x + 1) - f(2x) + f( {x}^{2} + 1) \\ = (4 {x}^{2} + 10x + 6) - (4 {x}^{2} + 6x + 2) + ( {x}^{4} + 5 {x}^{2} + 6) \\ = 4 {x}^{2} + 10x + 6 - 4 {x}^{2} - 6x - 2 + {x}^{4} + 5 {x}^{2} + 6 \\ = {x}^{4} + 5 {x}^{2} + 4x + 10[/tex]
Semoga dapat dipahami