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

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SUMMARY

The discussion focuses on finding inflection points for the function y = (x^2 - 7)e^x by analyzing its second derivative. The second derivative is calculated as e^x[(x - 1)(x + 5)] = 0, leading to potential inflection points at x = -∞, 1, and -5. However, it is established that x = -∞ is not a valid inflection point since it does not represent a finite value where the function's concavity changes. The valid inflection points are confirmed to be at x = 1 and x = -5.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and inflection points.
  • Familiarity with exponential functions, particularly e^x.
  • Knowledge of the behavior of functions in relation to convexity and concavity.
  • Ability to solve equations involving polynomials and exponential terms.
NEXT STEPS
  • Study the properties of exponential functions and their derivatives.
  • Learn how to determine concavity and convexity using the second derivative test.
  • Explore the concept of limits, particularly how they apply to infinity in calculus.
  • Practice finding inflection points for various functions to reinforce understanding.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and inflection points, as well as educators looking for examples of function behavior analysis.

knowLittle
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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 ?
 
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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
 

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