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

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

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Answer 1 · 2016-03-07T04:46:47

$1/e$

Explanation:

The instantaneous rate of change of a function $f(x)$ at a given point is the slope of the tangent line at that point, that is, the derivative $f'(x)$ evaluated at that point.

In this case, we have

$f'(x) = xe^x+e^x-x^2e^x-2xe^x$

$=-e^x(x^2+x-1)$

Then, evaluated at $x=-1$ we have

$f'(-1) = -e^(-1)((-1)^2+(-1)-1)$

$=-1/e(-1)$

$=1/e$

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

1 Answer

Explanation:

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

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