f = ax e^(bx^2)
f' = (a + ax(2bx)) e^(bx^2)
= a(1+2bx^2) e^(bx^2)
So, f achieves a max/min where f'=0
That is, where a(1+2bx^2) = 0
Or, where x^2 = -1/2b
So, it looks like we need b < 0, since x^2 is always positive
Not surprisingly, the value of a does not affect the presence of a max or min; it's just a scale factor. However, the sign of a will determine whether the extremum is a max or a min.
Anyway, we want f(1) to be a maximum. So, 1 = ±√(-1/2b), meaning b = -1/2, so we have
f(x) = ax e^(-x^2/2)
f(1) = a/√e, so
a = 7√e, and we have
f(x) = 7√e x e^(-x^2/2)
I'm sure you can prove that this has a max at x=1, and not a min. The graph shown here illustrates that we have a solution.
http://www.wolframalpha.com/input/?i=plot+y%3D7%E2%88%9Ae+x+e^%28-x^2%2F2%29%2Cy%3D7
Hello! I am taking Calculus I and am in Sec. 4.3 of Stewart's 7th Edition. I ran across a problem that I don't understand the mechanics of, could someone show me? Thanks a bunch!
For what values of the numbers a and b does the function
f(x)= axe^{b(x^2)}
have the maximum value f(1)=7?
2 answers
That was so helpful!! I appreciate you writing all that out, thank you very much!