Identify a local maxima for:
$y = x^{3} - 3 x + 2$

x = 2

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x = 1

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x = - 2

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x = - 1

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Solution

The correct option is C $x = - 1$
$y = f (x)$ cuts the $x$ -axis at $x = - 2$ and $x = 1$ .
Now $f^{'} (x)$ $= 3 x^{2} - 3$
$= 3 (x^{2} - 1)$ $> 0$
$x^{2} - 1 > 0$
$(x - 1) (x + 1) > 0$
$x < - 1$ and $x > 1$
Hence, it is an increasing function in $(- \infty - 1) \cup (1, \infty)$
And therefore it is a decreasing function in the interval $(- 1, 1)$ .
Now $f (1) = 0$ and $f (- 1) = - 1 + 3 + 2 = 4$
$f^{''} (x)_{x = - 1} < 0$
Hence, it attains a maximum at $x = - 1$ .

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