I keep getting this derivative wrong no matter how I approach the probelm. It uses constants which is throwing me off a little bit.

Question

The problem states : Let h(x)= xe^-ax^2 +10, where a is a constant.

a) for What values of a does the function h(x) have/has critical points.

b) At the valid Critical point, state is it is a local maximum or minimum.

C) State is the critical point is also a global maximum or a global minimum.

 

My best attempt.

I found the derivative for my first step. I got:

xe^-ax^2 * -2ax

I then tried to find the critical points by setting the derivative equal to zero but I keep getting error messages in my calculator. I'm not sure what I am doing wrong.

Then for C I thought I could take the second derivative and set that equal to zero (I also got errors when I did this) but the second derivative I got is xe^-ax^2 * -2a.

Please help me figure out how to resolve this problem. Much help is appreciated.

Answer 1

Here's my help for first derivative.

 

h'(x) = e^{-ax^2} + xe^{-ax^2}(-2ax) = e^{-ax^2}(1-2ax²) and setting this to zero to find critical points gets

x = ±1/√(2a).  (Assumes a>0)

 

 

Answer 2

This function domain is [-inf ,inf] and the range is depends on value of a if a<0 the domain will become  [0,inf] f a>0 then the domain will be [0 max value]

 

h'(x)=e(-a*x^2)-(2ax^2)*e^(x^2)=e^(-ax^2)*[1-2ax^2]=0

 

x*=+-1/sqrt(2a) and  inf You can understand the necessity of a>0

 

h''=-2*a*x*e^(-ax^2) [1-2ax^2]+e^(-ax^2)[-4ax]

=e^(-ax^2)[-6ax+4ax^3]

If you substitute the value of x*

h''(x)=[e^(-1/2)]*[-6*sqrt(a/2)+2*sqrt(a/2)]=)=[e^(-1/2)]*[-4*sqrt(a/2)]<0 i.e a>0

 

So if a>0 then this solution x*=1/sqrt(2a) is maximum

if you substitute x=inf both  h' and h''=0 so it is a saddle point.

More over if x=0 h(x) =0   so x=0 is the minimum point . If you take x*=-1/sqrt(2a) Just try yourself