The maximum value of $y=x^3-3 x$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 at x = -1
$y=x^3-3 x$
$y'=3x^2-3=0⇒x=±1$
$y''=6x$
$y''(-1)=-6<0$ (point of maxima)
$y_{max}=(-1)^3-3(-1)=2$
The maximum value of $y=x^3-3 x$ is:
Correct answer: option 2
The correct answer is Option (2) → 2 at x = -1
$y=x^3-3 x$
$y'=3x^2-3=0⇒x=±1$
$y''=6x$
$y''(-1)=-6<0$ (point of maxima)
$y_{max}=(-1)^3-3(-1)=2$