If #1/3(x-1)^3+2#, what are the points of inflection, concavity and critical points?

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Answer 1 · 2017-11-08T12:52:56

The point of inflection is #=(1,2)#. The interval of concavity is #x in (-oo,1)#. The interval of convexity is #(1,+oo)#

Let #f(x)=1/3(x-1)^3+2#

Let's develop #f(x)#

#f(x)=1/3(x^3-3x^2+3x-1)+2#

#=x^3/3-x^2+x+5/3#

Calculate the first derivative

#f'(x)=x^2-2x+1=(x-1)^2#

The critical points are when #f'(x)=0#

That is,

#(x-1)^2=0#, #=>#, #x=1#

Now, calculate the second derivative

#f''(x)=2x-2#

The points of inflections are when #f''(x)=0#

That is,

#2x-2=0#, #=>#, #x=1#

Make a sign chart

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,1)##color(white)(aaaa)##(1,+oo)#

#color(white)(aaaa)##Sign f''(x)##color(white)(aaaaa)##-##color(white)(aaaaaaaa)##+#

#color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu#

graph{1/3(x-1)^3+2 [-8.335, 9.445, -2.355, 6.534]}