Let XXX have CDF F(x)=1−(1+x)e−xF(x) = 1 - (1+x)e^{-x}F(x)=1−(1+x)e−x for x≥0x \ge 0x≥0. What is the PDF f(x)f(x)f(x)?
xe−xxe^{-x}xe−x
e−xe^{-x}e−x
(1+x)e−x(1+x)e^{-x}(1+x)e−x
x2e−xx^2e^{-x}x2e−x