# Concavity, inflection point, maxima, minima

• eMac
In summary, the conversation discusses finding the vertical and horizontal asymptotes, intervals of increase and decrease, local maximum and minimum values, intervals of concavity, and inflection points for the function f(x) = 1 / (L - x^1/2). The vertical asymptote was found to be x=0 and the horizontal asymptote to be y=0. The first derivative was used to find the intervals of increasing and decreasing values, as well as the minimum and maximum values. The second derivative was used to determine the concavity, and it was concluded that there are no inflection points. However, there seems to be some confusion about the function itself, as it is unclear if it is f(x) = 1

#### eMac

1.Find the vertical and horizontal asymptotes (if any), intervals of increase and decrease, local maximum and minimum values, intervals of concavity and inflection points for f(x) = 1 / (L - x^1/2)

3. I found the vertical asymptote to be "x=0" , I found the horizontal asymptote to be "y=0"
So to find intervals of increase and decrease you have to find the derivative, which I found to be 2*x/(L^2-2*x^2*L+x^4) from here I plugged in values.

To find local maximum and minimum values you have to find the second derivitive, which I found to be (2*L+6*x^2)/(L^3-3*x^2*L^2+3*x^4*L-x^6)

Then I came to the conclusion that there is no inflection point because there are no local maximum's meaning no change in concavity

I'm not sure if I did these right, would really appreciate some help, thanks.

Look over your methods of finding critical points and points of inflection.
What does the first derivative mean?

Yes, I think I stated it wrong in my first post I used the first derivative to find the intervals of increasing and decreasing values and also for minimum and maximum values. I used the second derivative to find the concavity.

Do you really mean f(x)=1/(L-sqrt(x))? Why do you think there is a vertical asymptote at x=0? And I don't think any of your derivatives are correct either. Or did you mean something else altogether?

No I meant L - x^2

eMac said:
No I meant L - x^2

Ok, so f(x)=1/(L-x^2), right? Still doesn't have a vertical asymptote at x=0. But at least your derivatives are making more sense. When you are thinking about this, I wouldn't expand powers of (L-x^2). Just leave them as they are.

When taking the derivative, remember L is not a variable, just a number.

## 1. What is concavity?

Concavity refers to the curvature of a graph or function. A graph is considered concave if it curves upward, and convex if it curves downward.

## 2. What is an inflection point?

An inflection point is a point on a graph where the curvature changes from concave to convex, or vice versa. It is also where the second derivative of a function is equal to zero.

## 3. How do you find the maximum and minimum points on a graph?

The maximum point on a graph is the highest point, or peak, while the minimum point is the lowest point, or valley. To find these points, you can take the first derivative of the function and set it equal to zero. Then, solve for the x-value(s) which will give you the coordinates of the maximum or minimum point.

## 4. What is the difference between a local and global maximum or minimum?

A local maximum or minimum is a point on a graph that is the highest or lowest within a small interval, but not necessarily the highest or lowest overall. A global maximum or minimum, on the other hand, is the highest or lowest point on the entire graph.

## 5. How do you determine the concavity and inflection points of a function?

To determine the concavity of a function, you can look at the sign of the second derivative. If it is positive, the function is concave upward, and if it is negative, the function is concave downward. To find the inflection points, you can set the second derivative equal to zero and solve for the x-value(s).