For what values of x is #f(x)=(x-3)(x-1)(x-2)# concave or convex?

1 Answer
Feb 24, 2018

#f(x)# is convex on the interval #(-∞,2)#; #f(x)# is concave on the interval #(2, ∞)#

Explanation:

Find the second derivative, #f''(x)#:

#f'(x)=(x-3)(x-1)d/dx(x-2)+(x-3)(d/dx(x-1))(x-2)+(d/dx(x-3))(x-1)(x-2)#

#f'(x)=(x-3)(x-1)+(x-3)(x-2)+(x-1)(x-2)#

#f''(x)=(x-3)(d/dx(x-1))+(x-1)(d/dx(x-3))+(x-3)((d/dx(x-2))+(x-2)(d/dx(x-3))+(x-1)(d/dx(x-2))+(x-2)((d/dx(x-1))#

#f''(x)=(x-3)+(x-1)+(x-3)+(x-2)+(x-1)+(x-2)#

#f''(x)=2(x-3)+2(x-1)+2(x-2)#

Set #f''(x)=0# and solve for #x:#

#2(x-3)+2(x-1)+2(x-2)=0#

#2x-6+2x-2+2x-4=0#

#6x-12=0#
#6x=12#
#x=12/6=2#

The domain of #f(x)# is (-∞,∞), as all polynomials are continuous. Let's break up the domain of #f(x)# around the value of #x# we've found:

#(-∞,2), (2,∞)#

Now, we must determine whether #f''(x)# is positive or negative in each of these intervals. If #f''(x)>0# in an interval, #f(x)# is concave on that interval. If #f''(x)<0# on an interval, #f(x)# is convex on that interval.

#(-∞,2):#

#f''(0)=2(-3)+2(-1)+2(-2)<0#

#f(x)# is convex on the interval #(-∞,2)#

#(2,∞):#

#f''(3)=2(0)+2(2)+2>0#

#f(x)# is concave on the interval #(2, ∞)#