Find maximum/minimum value of $y$ in $y = x^{3} - 3 x$

y_{m i n} = - 2

x = 1

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y_{m a x} = 2

x = - 1

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y_{m a x} = 2

x = 2

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y_{m i n} = - 2

x = - 2

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Solution

The correct options are
A $y_{m i n} = - 2$ at $x = 1$
B $y_{m a x} = 2$ at $x = - 1$
$y = x^{3} - 3 x y^{'} = 3 x^{2} - 3$
$3 x^{2} - 3 = 0 x = \pm 1$
$y "= 6 x$
Which is negative at $x = - 1$ so a maximum will occur at $x = - 1$
and is positive at $x = 1$ so a minimum will occur at $x = 1$
$y_{m i n} = - 2$ at $x = 1$
$y_{m a x} = 2$ at $x = - 1$

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