Expert: Alon Mandes - 5/19/2009

Question
For this problem:
f(x)=e^(-3.5x^2)

When I found the derivative I got: (e^(-3.5x^2))(-7x)
I don't know if that is right, however I can't figure out where these parts of it:

a) Find all critical numbers of f. If there are no critical values, enter 'NONE'.

b)Use interval notation to indicate where f(x) is increasing.

(C) Use interval notation to indicate where f(x) is decreasing.


(D) List the x-coordinates of all local maxima of f. If there are no local maxima, enter 'NONE'.

(E) Find the x-coordinates of all local minima of f. If there are no local minima, enter 'NONE'.

(F) Use interval notation to indicate where f(x) is concave up.

(G) Use interval notation to indicate where f(x) is concave down.


(H) List the x values of all inflection points of f. If there are no inflection points, enter 'NONE'.


For a lot of these problems I really need help on finding where it INC/DEC and where it is concave up/down. I especially get stuck on the concave up/down. For this problem I didn't know how to find the critical points, and if I can't find them I don't know its local max/min. Could you run me through this problem so I have an idea of what I NEED to be doing. Especially on the concave stuff. I know you do the 1st deriv test of the 2nd derivative of the first derivative, but I get lost somewhere in it. Thanks.

BROOKE

Answer
your derivative is right :
f(x)=e^(-3.5x²)
f'(x)=-7xe^(-3.5x²)
f''(x)=-7e^(-3.5x²)+49x²e^(-3.5x²)=[49x²-7]e^(-3.5x²)
We only have 1 extremum point at x=0 & it's a saddle point.
Interval of Increasing : f'(x)>0 :-7xe^(-3.5x²)>0 for all x<0
Interval of Decreasing : f'(x)<0 :-7xe^(-3.5x²)<0 for all x>0
Interval of Concave UP  :f''(x)<0:[49x²-7]e^(-3.5x²)<0
                        [49x²-7]<0 --> x>1/√7 or x<-1/√7  
Interval of Concave DOWN:f''(x)>0:[49x²-7]e^(-3.5x²)>0
                                 [49x²-7]>0 --> -1/√7<x<1/√7

Alon.