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

Question

Scarlett Allison · Accepted Answer

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

Aurora Alvarado · Answer

undefined To determine the concavity of $ f(x) = (x-3)(x-1)(x-2) $, we need to examine its second derivative. The second derivative, $ f''(x) $, indicates the concavity of the function. If $ f''(x) > 0 $ for a given interval, the function is convex on that interval. If $ f''(x) < 0 $, the function is concave on that interval. Taking the second derivative of $ f(x) $: $$ f'(x) = (x-1)(x-2) + (x-3)(x-2) + (x-3)(x-1) $$ $$ f''(x) = 2(x-2) + 2(x-3) + 2(x-1) $$ $$ f''(x) = 2x - 4 + 2x - 6 + 2x - 2 $$ $$ f''(x) = 6x - 12 $$ For concavity or convexity, set $ f''(x) = 0 $ and solve for $ x $: $$ 6x - 12 = 0 $$ $$ 6x = 12 $$ $$ x = 2 $$ So, $ f(x) $ changes concavity at $ x = 2 $. To determine the concavity or convexity on each interval, pick test points within those intervals and check the sign of $ f''(x) $. - For $ x < 2 $: Pick $ x = 0 $. $ f''(0) = 6(0) - 12 = -12 $, so $ f(x) $ is concave. - For $ x > 2 $: Pick $ x = 3 $. $ f''(3) = 6(3) - 12 = 6 $, so $ f(x) $ is convex. Therefore, $ f(x) = (x-3)(x-1)(x-2) $ is concave for $ x < 2 $ and convex for $ x > 2 $.