At what values of x does the graph of y = e + 2xe-" + x²e have a point of inflection B C x = -1 only x = -1 and x = 1 x= = -3-√√2 and x = −3+√2 x = 1 -√√2 and x = 1+ √2 -X

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Question:**

At what values of \( x \) does the graph of \( y = e^{-x} + 2xe^{-x} + x^2e^{-x} \) have a point of inflection?

**Options:**

(A) \( x = -1 \) only

(B) \( x = -1 \) and \( x = 1 \)

(C) \( x = -3 - \sqrt{2} \) and \( x = -3 + \sqrt{2} \)

(D) \( x = 1 - \sqrt{2} \) and \( x = 1 + \sqrt{2} \)
Transcribed Image Text:**Question:** At what values of \( x \) does the graph of \( y = e^{-x} + 2xe^{-x} + x^2e^{-x} \) have a point of inflection? **Options:** (A) \( x = -1 \) only (B) \( x = -1 \) and \( x = 1 \) (C) \( x = -3 - \sqrt{2} \) and \( x = -3 + \sqrt{2} \) (D) \( x = 1 - \sqrt{2} \) and \( x = 1 + \sqrt{2} \)
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Given Equation:-

y=e-x+2xe-x+x2e-x

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