What are the local extrema of f(x)= x^3 - 9x^2 + 19x - 3 ?

1 Answer
Feb 2, 2017

f(x)_max =(1.37, 8.71)
f(x)_min =(4.63, -8.71)

Explanation:

f(x)= x^3-9x^2+19x-3

f'(x) = 3x^2-18x+19

f''(x) = 6x-18

For local maxima or minima: f'(x) =0

Thus: 3x^2-18x+19 =0

Applying the quadratic formula:

x=(18+-sqrt(18^2-4xx3xx19))/6

x=(18+-sqrt96)/6

x=3+-2/3sqrt6

x~= 1.367 or 4.633

To test for local maximum or minimum:

f''(1.367) < 0 -> Local Maximum

f''(4.633) > 0 -> Local Minimum

f(1.367) ~= 8.71 Local Maximum
f(4.633) ~= -8.71 Local Minimum

These local extrema can be seen on the graph of f(x) below.

graph{ x^3-9x^2+19x-3 [-22.99, 22.65, -10.94, 11.87]}