- #1

Karol

- 1,380

- 22

## Homework Statement

Only 15

## Homework Equations

First derivative=maxima/minima/vertical tangent/rising/falling

Second derivative=points of inflection/concave upward-downward

## The Attempt at a Solution

$$x=y^3+3y^2+3y+2~\rightarrow~1=3(y^2+2y+1)y'$$

$$y'=\frac{1}{3(y+1)^2}>0,~y\neq (-1),~y=(-1)\rightarrow~x=-1$$

So the vertical tangent is at x=(-1)

The second derivative:

$$y"=(-2)\frac{y+1}{3(y^2+2y+1)(y+1)^2}=\left( -\frac{2}{3} \right)\frac{1}{(y+1)^3}$$

The answer, i don't know exactly how to interpret it, is (1,-1)