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What is the instantaneous rate of change of #f(x)=(x^2-3x)e^(x) # at #x=2 #?

Answer 1

Ezekiel Alexander

#5e^2#

The instantaneous rate of change when #x=2# can be found through computing #f'(2)#.

To find #f'(2)#, first find #f'(x)#.

Apply the rule of the product:

#f'(x)=e^xd/dx[x^2-3x]+(x^2-3x)d/dx[e^x]#

#f'(x)=e^x(2x-3)+e^x(x^2-3x)#

#f'(x)=e^x(2x+3+x^2-3x)#

#f'(x)=e^x(x^2-x+3)#

#f'(2)=e^2(2^2-2+3)#

#f'(2)=5e^2#

#f'(2)~~36.945#

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Answer 2

Scarlett Allison

To find the instantaneous rate of change of the function ( f(x) = (x^2 - 3x)e^x ) at ( x = 2 ), we need to find the derivative of the function and then evaluate it at ( x = 2 ).

First, let's find the derivative of ( f(x) ) using the product rule and the chain rule:

[ f'(x) = (2x - 3)e^x + (x^2 - 3x)e^x ]

Now, we'll evaluate ( f'(x) ) at ( x = 2 ):

[ f'(2) = (2(2) - 3)e^2 + ((2)^2 - 3(2))e^2 ] [ f'(2) = (4 - 3)e^2 + (4 - 6)e^2 ] [ f'(2) = (1)e^2 + (-2)e^2 ] [ f'(2) = e^2 - 2e^2 ] [ f'(2) = (1 - 2)e^2 ] [ f'(2) = -e^2 ]

So, the instantaneous rate of change of ( f(x) ) at ( x = 2 ) is ( -e^2 ).

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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.