For what values of x is #f(x)= -4x^3-x+12# concave or convex?

Question

Christopher Allers · Accepted Answer

Concave up in the interval #(-oo,0)#

Cocave down in the interval #(0, oo)# First determine f'(x) = #-12x^2 -1# and f"(x)= -24x The inflection point is at f"(x)=0 which gives x=0, which is a inflection point. Thus in the interval #(-oo, 0)# f"(x) would be >0 and in the interval #(0, +oo)# f"(x) < 0 . In the first case f(x) would be concave up and in the second case it would be concave down.

Paisley Johnson · Answer

undefined To determine where the function $ f(x) = -4x^3 - x + 12 $ is concave or convex, we need to find the second derivative $ f''(x) $ and analyze its sign: 1. Find the first derivative $ f'(x) $. $$ f'(x) = -12x^2 - 1 $$ 2. Find the second derivative $ f''(x) $. $$ f''(x) = -24x $$ 3. Analyze the sign of $ f''(x) $ to determine concavity/convexity: - If $ f''(x) > 0 $ for a certain interval, then the function is concave up (convex) on that interval. - If $ f''(x) < 0 $ for a certain interval, then the function is concave down on that interval. For $ f''(x) = -24x $: - $ f''(x) > 0 $ for $ x < 0 $, so $ f(x) $ is concave up (convex) on this interval. - $ f''(x) < 0 $ for $ x > 0 $, so $ f(x) $ is concave down on this interval. Therefore, $ f(x) $ is concave up (convex) for $ x < 0 $ and concave down for $ x > 0 $.