How do you find the local extrema for # (x^2)(e^-x) # from [-2,4]?

Answer 1

#PMIN[0,0],PMAX[-2,4e^2]#

We get #f'(x)=e^(-x)(2x-x^2)# #f''(x)=e^(-x)(x^2-4x+2)# solving
#f'(x)=0# we get #x=0# #f''(0)>0# #x=2# #f''(2)=e^(-2)(-2)<0#

since

#f(-2)=4e^2>4e^(-2)#

we get the Points as above.

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Answer 2

To find the local extrema of ( x^2 \cdot e^{-x} ) on the interval [-2, 4], you first need to find its critical points. Critical points occur where the derivative of the function is zero or undefined.

  1. Find the derivative of the function: [ f'(x) = 2x \cdot e^{-x} - x^2 \cdot e^{-x} ]

  2. Set the derivative equal to zero and solve for ( x ): [ 2x \cdot e^{-x} - x^2 \cdot e^{-x} = 0 ] [ x(2 - x) \cdot e^{-x} = 0 ]

This gives two critical points: ( x = 0 ) and ( x = 2 ).

  1. Check the endpoints of the interval [-2, 4], which are -2 and 4.

  2. Evaluate the function at these critical points and endpoints to find the local extrema.

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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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