Is #f(x) =(x-2)^3+x-3# concave or convex at #x=11#?

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Answer 1 · 2018-01-17T22:15:49

convex

#f(x) = (x-2)^3 + x - 3#

#(x-2)^3 = x^3 + (3*-2*x^2) + (3*4*x^1) + (-2)^3#

#= x^3 -6x^2 + 12x -8#

#(x-2)^3 + x - 3 = x^3 -6x^2 + 12x -8 + x-3 #

#= x^3-6x^2+13x-11#

#f(x) = x^3-6x^2+13x-11#

#f'(x) = 3x^2-12x + 13#

#f''(x) = 6x - 12#

substitute given #x-#value into #2#nd derivative:

#f''(x) = 6x - 12#

#x = 11#

#6x - 12 = 66-12 = 54#

at #x = 11, f''(x) >0#

when #f''(x) >0# for a certain #x-#value, the graph is convex at this point.