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

Question

Scarlett Allison · Accepted Answer

#f(x)=(x-3)(x+2)(3x-2)# #implies f(x)=(x^2-x-6)(3x-2)# #implies f(x)=3x^3-5x^2-4x+12# If #f(x)# is a function and #f''(x)# is the second derivative of the function then, #(i) f(x)# is concave if #f(x)<0# #(ii) f(x)# is convex if #f(x)>0# Here #f(x)=3x^3-5x^2-4x+12# is a function. Let #f'(x)# be the first derivative. #implies f'(x)=9x^2-10x-4# Let #f''(x)# be the second derivative. #implies f''(x)=18x-10# #f(x)# is concave if #f''(x)<0# #implies 18x-10<0# #implies 9x-5<0# #implies x<5/9# Hence, #f(x)# is concave for all values belonging to #(-oo,5/9)# #f(x)# is convex if #f''(x)>0#. #implies 18x-10>0# #implies 9x-5>0# #implies x>5/9# Hence, #f(x)# is convex for all values belonging to #(5/9,oo)#

Cameron Alexander · Answer

undefined To determine the concavity of the function $ f(x) = (x-3)(x+2)(3x-2) $, we need to find its second derivative and then determine its sign. 1. Find the first derivative $ f'(x) $: $$ f'(x) = (x+2)(3x-2) + (x-3)(3x-2) + (x-3)(x+2)(3) $$ 2. Simplify $ f'(x) $: $$ f'(x) = 3x^2 - 2x + 6x - 4 + 3x^2 - 9x - 6 + 3x^2 - 3x + 6x - 6 $$ $$ f'(x) = 9x^2 - 8x - 16 $$ 3. Find the second derivative $ f''(x) $: $$ f''(x) = 18x - 8 $$ Now, to determine concavity: If $ f''(x) > 0 $ for a given interval, $ f(x) $ is concave up (convex) on that interval. If $ f''(x) < 0 $ for a given interval, $ f(x) $ is concave down (concave) on that interval. To find the values of $ x $ where $ f(x) $ is concave or convex, set $ f''(x) $ to zero and solve for $ x $: $$ 18x - 8 = 0 $$ $$ 18x = 8 $$ $$ x = \frac{8}{18} $$ $$ x = \frac{4}{9} $$ Now, test the intervals: 1. For $ x < \frac{4}{9} $: $ f''(x) < 0 $ => $ f(x) $ is concave down. 2. For $ x > \frac{4}{9} $: $ f''(x) > 0 $ => $ f(x) $ is concave up. Thus, $ f(x) $ is concave down for $ x < \frac{4}{9} $ and concave up for $ x > \frac{4}{9} $.