Let e(x) be an even function, and let o(x) be an odd function, such that \(e(x) + o(x) = \frac{6}{x + 2} + x^2 + 2^x\)

for all real numbers x. Find o(1).

Guest Oct 2, 2019

#1**+1 **

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

.Rom Oct 2, 2019

#1**+1 **

Best Answer

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

Rom Oct 2, 2019