<p>Hi, I am not a smart one so Im asking for help. (lol obviously)</p>
<p>Let f be the function given by f(x)=2x(e^2x)
A. Find Lim(as X approaces -infinitive) f(x) and Lim (as X approaches infinitive) f(x)
B. Consider the family of functions defined by y=bx(e^bx), where b is a nonzero constant. Show that the absolute minimum value of bx(e^bx) is the same for all nonzero values of b</p>
<p>Help would be greatly appreciated.</p>
<p>A. negative infinity: 0, e^(2* negative infinity) =0
positive infinity: infinity it's infinity * infinity
B. y'=(b^2)x(e^bx)=0 x=0 draw a sign chart to confirm that the sign changes from - to +, which means that it is a local minimum. The negative infinity of the function is 0, and the positive infinity of the function is infinity. Thus this shows that the graph continues to increase. Thus the absolute minimum value of bx(e^bx) is 0.</p>
<p>B. Set u=bx, v=e^bx.
Then y=uv, and y' = uv' + u'v
= bx.(e^bx)(b) + b.e^bx
= b(e^bx)(bx+1)
y'=0 when bx+1 = 0 i.e. x= -1/b
Compute y'', plug in this value of x, and show that y'' > 0 at x=-1/b
i.e. x= -1/b is a local minimum for y
At x = -1/b, y= b(-1/b) (e^b(-1/b))
or y = (-1) e^(-1) for all non-zero values of b</p>
<p>^
also state that both the limit as x tends to -infinity and as x tends to+infinity are greater than -1/e</p>