Question

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

Answer 1

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

Explanation:

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)`