What is the instantaneous rate of change of $f(x)=(x^2-3x)e^(x)$ at $x=2$ ?

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Answer 1 · 2015-12-14T03:20:03

$5e^2$

The instantaneous rate of change when $x=2$ can be found through computing $f'(2)$ .

To find $f'(2)$ , first find $f'(x)$ .

$f'(x)=e^xd/dx[x^2-3x]+(x^2-3x)d/dx[e^x]$

$f'(x)=e^x(2x-3)+e^x(x^2-3x)$

$f'(x)=e^x(2x+3+x^2-3x)$

$f'(x)=e^x(x^2-x+3)$

$f'(2)=e^2(2^2-2+3)$

$f'(2)=5e^2$

$f'(2)~~36.945$

What is the instantaneous rate of change of f(x)=(x^2-3x)e^(x) f(x)=(x^2-3x)e^(x) at x=2 x=2 ?