What is the instantaneous rate of change of #f(x)=-(x+2)e^(x-1) # at #x=0 #?

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Answer 1 · 2016-04-13T19:57:38

I found: #-3/e#

You can evaluate the instantaneous rate of change at a point evaluating the derivative of your function at that specific point.

We evaluate the derivative first:

#f`(x)=-e^(x-1)-(x+2)e^(x-1)#

using the Product and Chain Rule.

Now we calculate the derivative at #x=0# to get:

#f`(0)=-e^-1-2e^-1=-3/e#

that will be your instantanous rate of change.