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.

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