Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 32

min/max
Potential local extrema are found at points where y'=0.
y'=(2-2x)e^(2x-x^2)
(2-2x)e^(2x-x^2)=0
Since e^(2x-x^2)>0 we have
2-2x=0
x=1
To determine whether the function has minimum, maximum or extreme at all at the point x=1 we need to check the sign of derivative before and after that point.
y'(0)=(2-0)e^0=2>0=> function is increasing
y'(2)=(2-4)e^(4-4)=-2e^0=-2<0=> function is decreasing
Therefore, the function has local maximum at x=1. The value of the maximum is y(1)=e^(2-1)=e.
Points of inflection
Points of inflection are found where y''=0.
y''=-2e^(2x-x^2)+(2-2x)^2e^(2x-x^2)=e^(2x-x^2)(4x^2-8x+2)
e^(2x-x^2)(4x^2-8x+2)=0
4x^2-8x+2=0
2x^2-4x+1=0
Now we apply quadratic formula.
x_(1,2)=(4pm sqrt(16-8))/4=(4pm2sqrt2)/4=1pm sqrt2/2
x_1=1-sqrt2/2 x_2=1+sqrt2/2 Asymptotes
Since the domain of the function is set of all real numbers the function has no vertical asymptotes.
Horizontal asymptotes
lim_(x->oo)e^(2x-x^2)=e^(-oo)=0
lim_(x->-oo)e^(2x-x^2)=e^(-oo)=0
Since the function has horizontal asymptote for both -oo and oo it cannot have oblique asymptote.
Sketch of the curve can be seen in the image below.