How do you find the maximum, minimum and inflection points and concavity for the function #y = xe^x#?

Answer 1
  • #x=-1# is a local minimum and the absolute minimum of the function #y=xe^x# because #y'# changes from negative to positive at #x=-1#.
  • #x=-2# is an inflection point of the function #y=xe^x# because #y''# changes from negative to positive at #x=-2#.
  • #y=xe^x# is concave down (convex) on #x in (-oo,-2)#, and concave up on #x in (-2,oo)#
#y=(x)(e^x)#
To find maxima and minima, find where #y'# is equal to zero: #y'=(x)(e^x)+(1)(e^x)# #y'=(e^x)(x+1)# #0=(e^x)(x+1)# #0=x+1# #x=-1# To check whether #x=-1# is a max or a min, use the first derivative test (plug in values less than and greater than -1): #y'(-100)="negative"# #y'(100)="positive"# #x=-1# is a local minima and the absolute minimum of the function #y=xe^x# because #y'# changes from negative to positive at #x=-1#.
Do the second derivative test to find inflection points and concavity: #y'=(e^x)(x+1)# #y''=(e^x)(1)+(e^x)(x+1)# #y''=(e^x)(x+2)# #0=(e^x)(x+2)# #0=x+2# #x=-2# #y''(-100)="negative"# #y''(100)="positive"# #x=-2# is an inflection point of the function #y=xe^x# because #y''# changes from negative to positive at #x=-2#. #y=xe^x# is concave down (convex) on #x in (-oo,-2)#, and concave up on #x in (-2,oo)#
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Answer 2

To find the maximum and minimum points of the function ( y = xe^x ), we first take the derivative of the function and set it equal to zero to find critical points. Then we use the second derivative test to determine the nature of these critical points.

  1. Derivative of ( y = xe^x ): ( y' = e^x + xe^x ).
  2. Set ( y' = 0 ): ( e^x + xe^x = 0 ).
  3. Solve for ( x ): ( x = -1 ) (only real solution).

To find inflection points and concavity, we follow these steps:

  1. Second derivative of ( y = xe^x ): ( y'' = e^x + xe^x ).
  2. Substitute ( x = -1 ) into ( y'' ): ( y''(-1) = e^{-1} - e^{-1} = 0 ).

Since the second derivative is zero at ( x = -1 ), we have an inflection point at this value. To determine concavity, we examine the behavior of the second derivative around ( x = -1 ):

  • If ( y'' ) changes sign from negative to positive at ( x = -1 ), the function is concave up.
  • If ( y'' ) changes sign from positive to negative at ( x = -1 ), the function is concave down.

Since ( y''(-1) = 0 ) and the sign changes from negative to positive as ( x ) increases past ( -1 ), the function ( y = xe^x ) is concave up for ( x > -1 ) and concave down for ( x < -1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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