# Verify if the function y=x^3+3x^2-3x+6 has local extrema?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A function has extreme values at the points where the first derivative is equal to 0.

y=x^3+3x^2-3x+6

y'= 3x^2 + 6x - 3

3x^2 + 6x - 3 = 0

=> x^2 + 2x - 1 = 0

x1 = -2/2 + sqrt (4 + 4)/2

=> -1 + 2*sqrt 2/2

=> -1 + sqrt 2

x2 = -1 - sqrt 2

The function has extremes at the points where x = -1 + sqrt 2 and x = -1 - sqrt 2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the local extrema of a function, we must calculate the first derivative zeroes.

We'll diiferentiate with respect to x:

dy/dx = (x^3+3x^2-3x+6)'

dy/dx = 3x^2 + 6x - 3

We'll cancel dy/dx = 0:

3x^2 + 6x - 3 = 0

We'll divide by 3:

x^2 + 2x - 1 = 0

We'll determine the zeroes of the quadratic:

x1 = [-2+sqrt(4 + 4)]/2

x1 = (-2+2sqrt2)/2

x1 = -1 + sqrt2

x2 = -1 - sqrt2

Since the function has critical points x1 and x2 and the local extrema f(x1) and f(x2).

Since the function is decreasing between x1 and x2 and it is increasing over the ranges (-infinite, -1-sqrt2) and (-1+sqrt2 , +infinite), then f(-1-sqrt2) is a maximum point and f(-1+sqrt2) is a minimum point.