Find Intervals, where Function is Convex or Concave and Inflection Points

knowLittle
Messages
307
Reaction score
3

Homework Statement


y= (x^2 -7) e^x

The Attempt at a Solution



I'm trying to find inflection points by setting the second derivative=0
I found that the derivative is:
##2xe^{x}+x^{2}e^{x}-7e^{x}=0##
##e^{x}[2x+x^{2}-7]=0##
Then, the 2nd derivative:
##e^{x}[(x-1)(x+5)]=0##, then the inflection points are at x=- infinity; 1; -5.
Where ##e^{x}=0 ##, happens when x=-infinity
Is it correct to use - infinity as a value of x ?
 
Last edited:
Physics news on Phys.org
knowLittle said:

Homework Statement


y= (x^2 -7) e^x





The Attempt at a Solution



I'm trying to find inflection points by setting the second derivative=0
I found that the derivative is:
##2xe^{x}+x^{2}e^{x}-7e^{x}=0##
##e^{x}[2x+x^{2}-7]=0##
Then, the 2nd derivative:
##e^{x}[(x-1)(x+5)]=0##, then the inflection points are at x=- infinity; 1; -5.
Where ##e^{x}=0 ##, happens when x=-infinity
Is it correct to use - infinity as a value of x ?

No, it's not. Basically, an inflection point x0 is a point where the behavior of f(x) changes from convex to concave (or opposite) as x increases through x0. How do you increase from values < -∞ to > -∞ (that is, how do you pass from one side of -∞ to the other)?

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top