Question

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

Answer 1

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

Explanation:

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]}