Question

Consider the family of functions by y=bxe^(bx), where b is a nonzero constant. Show that the absolute minimum value of bxe^(bx) is the same for all nonzero values of b.

Answer 1

You have to differentiate f:\[y'=be^{bx}+ bx \cdot be^{bx}=b(1+bx)e^{bx}\]
Then you should try to solve y'= 0.

Answer 2

Solving y'=0: sinc b is nonzero, and e^{bx} is always positive, y'=0 means:

1+bx=0, so x = -1/b.

Now if x < -1/b then y' < 0 and if x > -1/b, y'>0, so y has a minimum for x = -1/b.
Also, because the domain is all real numbers, there are no extremes on the edges of the domain, so we've found the x value of the abs. min.