Is $f(x)=e^x(x^2-x)$ increasing or decreasing at $x=3$ ?

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Answer 1 · 2018-05-12T14:04:01

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please follow the picture and i pointed 2 in the graph my mistake which should be x=3 .

Answer 2 · 2018-05-12T14:09:58

The function is increasing at $x=3$

Calculate $f'(x)$ and lok at the sign of $f'(3)$

$f(x)=e^x(x^2-x)$

The derivative is calculated with the product rule.

$(uv)'=u'v+uv'$

$u=e^x$ , $=>$ , $u'=e^x$

$v=x^2-x$ , $=>$ , $v'=2x-1$

Therefore,

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

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

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

When $x=3$

$f'(3)=e^3(3^2+3-1)=11e^3$

$f'(3)>0$

The function is increasing at $x=3$

graph{e^x(x^2-x) [-6.24, 6.244, -3.12, 3.12]}

Is f(x)=e^x(x^2-x)f(x)=e^x(x^2-x) increasing or decreasing at x=3x=3?