What are the local extrema of $f (x) = x^{3} - 3 x^{2} - 9 x + 1$ $f(x) = x^3 - 3x^2 - 9x +1$ ?

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Answer 1 · 2017-05-28T20:23:54

relative maximum: $(- 1, 6)$ $(-1, 6)$
relative minimum: $(3, - 26)$ $(3, -26)$

Given: $f (x) = x^{3} - 3 x^{2} - 9 x + 1$ $f(x) = x^3 - 3x^2 - 9x + 1$

Find the critical numbers by finding the first derivative and setting it equal to zero:

$f'(x) = 3x^2 -6x - 9 = 0$

Factor: $(3x + 3 )(x -3) = 0$

Critical numbers: $x = -1, " "x = 3$

Use the second derivative test to find out if these critical numbers are relative maximums or relative minimums:

$f''(x) = 6x - 6$

$f''(-1) = -12 < 0 => " relative max at " x = -1$

$f''(3) =12 > 0 => " relative min at " x =3$

$f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 1 = 6$

$f(3) =3^3 - 3(3)^2 - 9(3) + 1 = -26$

relative maximum: $(-1, 6)$
relative minimum: $(3, -26)$

What are the local extrema of f(x) = x^3 - 3x^2 - 9x +1f(x)=x3−3x2−9x+1f(x) = x^3 - 3x^2 - 9x +1?