What are the local extrema, if any, of $f(x)= 2x^3 – 6x^2 – 48x + 24$ ?

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Answer 1 · 2016-07-02T12:09:02

local max at x = -2

local min at x = 4

$f(x)= 2x^3 – 6x^2 – 48x + 24$

$f'(x)= 6x^2 – 12x – 48 = 6(x^2 - 2x - 8)$

$= 6 (x-4) (x+2)$

$implies f' = 0$ when $x = -2, 4$

$f'' = 12(x - 1)$

$f''(-2) = -36 < 0$ ie max
$f''(4) = 36 > 0$ ie min

the global max min are driven by the dominant $x^3$ term so $lim_{x to pm oo} f(x) = pm oo$

it must look like this..

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What are the local extrema, if any, of f(x)= 2x^3 – 6x^2 – 48x + 24f(x)= 2x^3 – 6x^2 – 48x + 24?