How do use the first derivative test to determine the local extrema #f(x)=x^4-4x^3+4x^2+6 #?

1 Answer
Aug 8, 2015

Answer:

You get: 2 minima and 1 maximum.

Explanation:

We can study the derivative of the function:
#f'(x)=4x^3-12x^2+8x#
Set the derivative equal to zero:
#f'(x)=0#
#4x^3-12x^2+8x=0#
#4x(x^2-3x+2)=0#
With solutions:
#x_1=0#
#x^2-3x+2=0# using the Quadratic Formula:
#x_(2,3)=(3+-sqrt(9-8))/2=(3+-1)/2#
two solutions:
#x_2=2#
#x_3=1#
We can now study the sign of the derivative setting: #f'(x)>0# getting:
enter image source here
Graphically:
graph{x^4-4x^3+4x^2+6 [-20.27, 20.28, -10.14, 10.13]}