Course:MATH102/Question Challenge/2002 December Q2(iv)

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Question

Critical points of $y=x^{3}e^{-x}$ occur at:

(a) $x=0,\pm 3$

(b) $x=0,\pm 1/3$

(c) $x=0,1/3$

(d) $x=0,3$

(e) none of the above

${\displaystyle }$

Hints

[show]Hint
Remember your product rule :)

Solutions

[show]Solution
$y=x^{3}e^{-x}$ To find the critical points of the function, we take the first derivative and set it to zero: $dy/dx=3x^{2}e^{-x}-x^{3}e^{-x}=x^{2}e^{-x}(3-x)=0$ In this factored form, we see that there are two solutions, $x=0,3$ (Note that $e^{-x}>0$ for all $x$.) Therefore our critical points are: $x=0,3$ Answer is D correct