What is the instantaneous rate of change of f(x)=xex−(x+2)ex−1 at x=0?

Question

juniorvtqanqtdd · Accepted Answer

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} + x e^{x} - e^{x - 1} - (x + 2) e^{x - 1}

Let us evaluate it at x=0

f^{'} (0) = e^{0} + 0 e^{0} - e^{0 - 1} - (0 + 2) e^{0 - 1} = 1 - \frac{1}{e} - \frac{2}{e} = \frac{e - 3}{e} = - 0.103

PCCNQN4XKhjx · Accepted Answer

xex−(x+2)ex−1 and evaluate it at x=0.So, find the derivative of the function: ddx(xex−(x+2)ex−1) and evaluate it at x=0.So, find the derivative of the function: ddx(xex−(x+2)ex−1)=(−x+ex−3+e)ex−1Finally, evaluate the derivative at x=0.(ddx(xex−(x+2)ex−1))|{(x=0)}=((−x+ex−3+e)ex−1)|(x=0)=−3+eeTherefore, the instantaneous rate of change of f(x)=xex−(x+2)ex−1 at x=0 is −3+ee.

What is the instantaneous rate of change of

Answered question

Answer & Explanation

New Questions in Calculus 1