Graphing Derivatives: How to Find Maxima, Minima, and Points of Inflection

[Karol]

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)

 

[Ray Vickson]

Homework Statement

View attachment 212347
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)

For $y = -1$ we have $x = +1$, not $x = -1$.

In the region $y > 0$ the function $f(y) = y^3 + 3 y^2 + 3y + 2$ is (strictly) convex, but for $y < 0$ it has mixed behavior: strictly convex for $y >-1$ and strictly concave for $y < -1$. Here, the modern terminology "convex" is what older introductory calculus books call "concave up", while "concave" is new-speak for "concave down".

The value $y=-1$ is a stationary point of $f(y)$ (derivative = 0), and is also an inflection point (second derivative = 0).

See, eg., https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cv1/t
or
http://mathworld.wolfram.com/ConvexFunction.html
regarding terminology.

 

[haruspex]

Homework Statement

View attachment 212347
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)

I assume that in taking the second derivative you are trying to determine whether the arc lies to the left or the right of the tangent.
You might find it easier to swap x and y and ask whether it is a local maximum or local minimum. How do you resolvethat when the second derivative is zero?

 

[Karol]

$$f(y) = y^3 + 3 y^2 + 3y + 2~\rightarrow~y'=3(y+1)^2,~y'=0~\rightarrow~x=(-1)$$
$$y(-1)~\rightarrow~x=(-1)$$
The y'=0 line that is parallel to the y-axis is vertical to the x.
The graph is:

 

[Karol]

Correction:

 

[haruspex]

Correction:
View attachment 212432

Right... so do you still have a question?

 

[Karol]

No, thank you Ray and Haruspex

 

[epenguin]

I think your calculations might have been easier and your result more obvious to you, related perhaps to something already familiar or known, perhaps all visible in a trice rather than the forgettable result of a slog if you had rewritten 15 as
$$x=\left( y+1\right) ^{3}+1$$
Nowhere is there any law btw forcing people to always portray x as horizontal and y as vertical!