f(x) is defined for all real x, so its domain is R

f(-x)=(-x)e^(x)(b)Is f(x) an odd or even orneither

which is not equal to either f(x), or -f(x), so f is neither odd nor even

neither x nor e^(-x) has a discontinuity in R so neither does their(c)determine any discontunity point for f(x)

product.

L'hopital's rule is needed here:(d)evaluate lim (x->infinity) f(x) and lim (x->-infinity) f(x)

f(x)=x e(-x) = x/e(x)

so:

lim(x->inf) f(x) = (lim(x->inf) d/dx(x) )/(lim(x->inf) d/dx(e^x)) =1/(lim(x->inf) e^x)=0

and

lim(x->-inf) f(x) = (lim(x->-inf) d/dx(x) )/(lim(x->-inf) d/dx(e^x)) =1/(lim(x->-inf) e^x)=inf

(d/dx) x e^(-x)=e^(-x) + x (-1) e^(-x)=e^(-x)[1-x]e)find critical numbers and where the function is increasing or decreasing

thus the critical points are the solutions of:

e^(-x)[1-x]=0,

but e^(-x) is never zero., so we want the solution/s of 1-x=0, which is

of course x=1.

For x large positive df/dt is negative, so f is decreasing on (1,inf), and so

increasing on (-inf,1).

RonL