(1)f(x)=xe2x(a)f(0)=∞,Noy−intercept(b)vertical assumptote does not exist sinceex/2=0,x→−∞f(∞)=0Horizontal assumptote isy=0.(c)f′(x)=21xe2x−x2e2xf(x)rises iff′(x)>0⟹2x2xe2x−2e2x>0⟹e2x(x−2)>0⟹x>2f(x)fallsf′(x)<0⟹2x2xe2x−2e2x<0⟹e2x(x−2)<0⟹x<2Local extremaf(2)=2e(d)f(x)is concave up, ifff′′(x)>0⟹f(x)=4x42x2e2x+x3e2x−2x2e2x−4x(x−2)e2x>0⟹x2−4x+8>0The roots of the quadraticequation are complexHence, the function is not concave upf(x)is concave down, ifff′′(x)<0⟹f(x)=4x42x2e2x+x3e2x−2x2e2x−4x(x−2)e2x<0⟹x2−4x+8<0The roots of the quadraticequation are complexHence, the function is not concave downThere is inflexion point ifff′′(x)=0But, since the roots of thequadratic equation are complex,There are no inflexion points(2)Area of the rectangle=xyVolume of cylinder=πr2hA larger surface area gives alarger volume.2(x+y)=36,x+y=18A=x(18−x)A=x(18−x)A′(x)=18−2xAis maximum or minimum atA′(x)=0⟹18−2x=0,x=9y=18−x=18−9=9x=9cmandy=9cmgives the largest volume.
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