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

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Answer 1 · 2017-05-28T20:23:54

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

Given: #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)#