The interval in which $f(x)=x^2 e^{-x}+7$ is increasing is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $(0,2)$
$y=x^2 e^{-x}+7$
$y'=2xe^{-x}-x^2e^{-x}=0$
$e^{-x}x(2-x)=0$
$⇒x=0,2$
f(x) increasing in (0, 2)
The interval in which $f(x)=x^2 e^{-x}+7$ is increasing is:
Correct answer: option 4
The correct answer is Option (4) → $(0,2)$
$y=x^2 e^{-x}+7$
$y'=2xe^{-x}-x^2e^{-x}=0$
$e^{-x}x(2-x)=0$
$⇒x=0,2$
f(x) increasing in (0, 2)