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

• Karol
I do not think I ever recall seeing a book or article where the x-axis was vertical and the y-axis was horizontal, but that might just be a matter of happenstance. ]
Karol

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)

Karol said:

## 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.

Karol said:

## 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?

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

Correction:

No, thank you Ray and Haruspex

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!

Last edited:

## 1. What is the purpose of drawing graphs by derivatives?

The purpose of drawing graphs by derivatives is to visualize the relationship between a function and its derivative. This can help to understand the behavior of the function and its rate of change at different points.

## 2. How do you draw a graph by using derivatives?

To draw a graph using derivatives, you first need to find the derivative of the original function. Then, plot the points of the derivative on a graph, using the original function's input values as the x-coordinates. Finally, connect the points to create a smooth curve.

## 3. What information can be obtained from a graph drawn by derivatives?

A graph drawn by derivatives can provide information about the slope, concavity, and critical points of a function. It can also show the rate of change of the function at different points.

## 4. Can a graph drawn by derivatives be used to find the maximum and minimum points of a function?

Yes, a graph drawn by derivatives can be used to find the maximum and minimum points of a function. The maximum and minimum points can be identified as the points where the graph of the derivative crosses the x-axis.

## 5. Are there any limitations to drawing graphs by derivatives?

One limitation of drawing graphs by derivatives is that it only works for functions that are differentiable. Functions with discontinuities or sharp corners cannot be accurately represented by their derivatives. Additionally, the graph may not accurately represent the behavior of the function near its endpoints.

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