Question

f(x)= 2x^3-6x+1 demonstrati ca f(2012)+f(2014)<\= decat f(2013)+f(2015) ajutorrrr va rogg

Answer 1

$f(x)= 2x^3-6x+1 \\ \\ f(x)+f(x+2) = 2x^3-6x+1+2(x+2)^3-6(x+2)+1 = \\ \\ =2x^3-6x+1+2(x+2)\Big((x+2)^2-3\Big)+1 = \\ \\ = 2x^3-6x+2(x+2)(x^2+4x+1)+2 = \\ \\ = 2x^3-6x+2\cdot (x^3+4x^2+x+2x^2+8x+2)+2 = \\ \\ = 2x^3-6x+2x^3+12x^2+18x+6 = \\ \\ =4x^3+12x^2+12x+6 \\ \\ \Rightarrow f(x)+f(x+2) = 4x^3+12x^2+12x+6\\ \\ \\ f(2012)+f(2014)\leq f(2013)+f(2015) \\ \\ f(2012)+f(2012+2) \leq f(2013)+f(2013+2)$

[tex]4\cdot (2012)^3+12\cdot (2012)^2+12\cdot 2012+6 \leq \\ \leq 4\cdot (2013)^3+12\cdot (2013)^2+12\cdot 2013+6\Big|:4 \\ \\ 2012^3+3\cdot 2012^2+3\cdot 2012 \leq 2013^3+3\cdot 2013^2+3\cdot 2013 \\ \\ \left\{ \begin{array}{ll} 2012^3\leq 2013^3 \\ 3\cdot 2012^2\leq 3\cdot 2013^3 \\ 3\cdot 2012\leq 3\cdot 2013 \end{array} \right| \\ ~~~~\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_~(+) \\ \\ 2012^3+3\cdot 2012^2+3\cdot 2012 \leq 2013^3+3\cdot 2013^2+3\cdot 2013 \quad $(A)$ [/tex]


$\Rightarrow f(2012)+f(2014)\leq f(2013)+f(2015) \quad (A)$