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

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

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Answer 1 · 2017-02-21T13:46:35

I got: $(e-3)/e=-0.10364$

Explanation:

To evaluate the instantaneous rate of change we need to evaluate the Derivative of our function and evaluate it at $x=0$ .
So we get:
$f'(x)=e^x+xe^x-e^(x-1)-(x+2)e^(x-1)$

let us evaluate it at $x=0$
$f'(0)=e^0+0e^0-e^(0-1)-(0+2)e^(0-1)=1-1/e-2/e=(e-3)/e=-0.10364$

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

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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