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

Question

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

Paisley Johnson · Accepted Answer

Concave (Convex) Up on the interval #( -oo,1 )#
Concave (Convex) Down on the interval #( 1, oo )# We are given the function #f(x) = -x^3 + 3x^2 - 2x + 2# #color(red)(Step.1)# Find the First Derivative #f'(x) = -3x^2 + 6x -2# #color(red)(Step.2)# Find the Second Derivative #f''(x) = -6x + 6# #color(red)(Step.3)# Next, set #f''(x) = -6x + 6 = 0# Simplifying, we get #x = 1# #color(red)(Step.4)# Then, we consider a number larger than 1 and a number smaller than 1 and substitute the values in our Second Derivative. If the number is Greater than 1, our #f''(x) = -6x + 6"# will yield a "Negative" number. If the number is Less than than 1, our #f''(x) = -6x + 6"# will yield a "Positive" number. Hence, we observe that #f(x)# is "Concave Up" on the interval #(-oo, 1)# and "Concave Down" on the interval #(1, oo)# Refer to the Number Line as shown below: :..........................................................*..................................................................:

Scarlett Allison · Answer

undefined To determine the concavity of $ f(x) = -x^3 + 3x^2 - 2x + 2 $, we need to find its second derivative, $ f''(x) $, and then analyze its sign. First derivative: $ f'(x) = -3x^2 + 6x - 2 $ Second derivative: $ f''(x) = -6x + 6 $ $ f''(x) $ is negative when $ x < 1 $ and positive when $ x > 1 $. Therefore, the function is concave down for $ x < 1 $ and concave up for $ x > 1 $.