Exponential Functions?

Question

Determine the 1st derivative of g(x) = 2xe^-x2.
Find the maximum value of the function g(x) = 2xe^-x2.
Also note that e^a is always positive for all a

Answer 1

The first derivative of
2xe^-(x^2) is
2e^(-x^2)- 4x^2(e^(-x^2)) using product rule
this simplifies to
e^-(x^2) [2-4x^2]

the maximum value of the function g(x) is found by setting
the second derivative equal to zero

so
we can set
(2-4x^2) = 0
because anything times zero is zero
so e^(-x^2)(0) = 0 so solving
(2-4x^2) = 0
solving the quadratic gives the following solutions
x = 0.707 and x = -0.707
plut x = 0.707 into g(x) gives
0.858 and plugging x = -0.707 into g(x) gives
-0.858
the maximum value of g(x) occurs at x = 0.707 and y = 0.858

Answer 2

I am taking x2 to be x squared and I am writing it as x*x
and e^ as exp.

dg/dx = 2 * [exp(-x*x)*d/dx(x) + x*d/dx[exp(-x*x)]

dg/dx = 2*(exp(-x*x) + x(-2x)exp(-x*x)
dg/dx = 2*exp(-x*x)*(1 - 2x*x)

set dg/dx = 0 for the maximum so 1 - 2x*x = 0 and x=SQRT(2)/2

Since SQRT(2)/2 is about 0.707, the following table shows this answer to be correct.

0.1 0.198009967
0.2 0.384315776
0.3 0.548358711
0.4 0.681715031
0.5 0.778800783
0.6 0.837211591
0.7 0.857676952
0.8 0.843667878
0.9 0.800744519
1 0.735758882
2 0.073262556
3 0.000740459
4 9.00281E-07
5 1.38879E-10
6 2.78343E-15
7 7.34004E-21
8 2.5661E-27
9 1.19514E-34

Answer 3

its
[2x[-2x]e^[-x2]]+[2e^[-x2]]
when we equate this to 0
we get e^[-2x]=0 and 2x^2=1
thus the only answer is from 2x^2=1
ie x^2=.5 and
x^2= -.5 which is not possible
thus x=sqrt[.5] is the only answer