How do use the first derivative test to determine the local extrema #f(x) = (x+1)(x-3)^2#?

1 Answer
Aug 31, 2015

Answer:

Have a look:

Explanation:

You can use the Product and Chain Rule to find the derivative:
#f'(x)=1*(x-3)^2+(x+1)*2(x-3)=x^2-6x+9+2(x^2-3x+x-3)=#
#=x^2-6x+9+2(x^2-2x-3)=x^2-6x+9+2x^2-4x-6=#
#=3x^2-10x+3#
Now you can set it equal to zero to find the x coordinate(s) of the local extrema:
#3x^2-10x+3=0#
We can use the Quadratic Formula:
#x_(1,2)=(10+-sqrt(100-36))/6=(10+-8)/6=#
So you get:
#x_1=3# giving #y=0#
#x_2=1/3# giving #y=9.5#

Graphically:
graph{(x+1)((x-3)^2) [-10, 10, -5, 5]}