derivative of e?

Question

ive got e^(-x^2) and f'(x) = e^(-x^2)-2x using the chain rule. But then how would I find what x is to get f'(x) to equal 0?

Answer 1

f(x) = exp(-x^2)

then

f'(x) = exp(-x^2) * (-2x) <---- chain rule

f'(x) = 0
exp(-x^2) * (-2x) = 0
exp(-x^2) = 0 OR (-2x) = 0
[no soultion] OR x = 0

The only solution is x = 0.


Note: the graph of f is a "bell curve" as you often encounter in statistics. You know that that curve has only one point where the tangent is horizontal.

Answer 2

If -2x*e^(-x^2) = 0, then either -2x = 0 or e^(-x^2)= 0.

So x= 0 from the first equation. The second eqn. has no solution, because e to any power is always positive. (SInce e is a positive constant itself)

Answer 3

the easiest way to answer this is to use a math tool like maple or matlab and graph the function. the point that the function crosses the X axis would be your answer. you would need numerical methods to solve functions like this. ( i am assuming that you did the derivation properly)

Answer 4

This site will clear up all of your answers! The derivative of e^x is e^x.......http://mathforum.org/library/drmath/view/60705.html

Answer 5

x^2 = t proceed

Answer 6

derivative of e^x is e^x