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

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Answer 1 · 2015-12-14T03:20:03

#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)#.

#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#