Calculus graph element spotting

In summary: The curve is concave "good" which is to say concave upward on the left and downward on the right. At x2 the concavity changes- so there is an inflection point there.
  • #1
Blade
12
0
http://www.delteria.net/images/calctest.gif

Conditions:
a. f'(x)=0
b. f"(x)= 0
c. f'(x) = DNE
d. f = relative max
e. f = point of inflection

What I have so far (they can repeat I believe):
a. X0, X4
b. X3
c. X1
d. X2
e. X2

I'm sure something is wrong... Also, what would a f"=0 look like?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
This really should be under "homework".

a) Yes, at x0 and x4, the tangent line is horzontal so f'= 0.

b) f"= 0 means the curve has 0 "curvature"(!) and so is very "straight" at least for a short distance. I would agree that it looks like the curve is very straight at x3 but I recommend you also look closely at x2. f"> 0 means the curve is "concave" upward while f"< 0 means it is concave downward. f"= 0 where the concavity changes.

c) Yes, there is a "cusp" at x1 and so the derivative does not exist.

d) "relative max" should be easiest of all but it surely doesn't happen at x2! Forget about derivatives and just ask yourself "where does the curve to up to the point and then back down again?"

e) A "point of inflection" is where the second derivative exists but changes sign (and so must be 0). Look at (b) again.
 
  • #3


Your response is mostly correct. Here are a few corrections and clarifications:

a. f'(x)=0: This condition means that the slope of the function at that point is 0. In the provided graph, this occurs at X0 and X4. So your response is correct.

b. f"(x)=0: This condition means that the second derivative of the function is 0. In the provided graph, this occurs at X3. So your response is correct.

c. f'(x) = DNE: This condition means that the derivative of the function does not exist at that point. In the provided graph, this occurs at X1. So your response is correct.

d. f = relative max: This condition means that the function has a maximum value at that point. In the provided graph, this occurs at X2. So your response is correct.

e. f = point of inflection: This condition means that the concavity of the function changes at that point. In the provided graph, this occurs at X2. So your response is correct.

For f''=0, it would look similar to b. f"(x)=0, except the point would be labeled as X2 instead of X3. This is because the second derivative is 0 at X2, indicating that the concavity changes at that point.
 

Related to Calculus graph element spotting

1. What is "Calculus graph element spotting"?

"Calculus graph element spotting" is the process of identifying and analyzing different elements present in a graph of a calculus equation, such as the behavior of the graph, critical points, and concavity.

2. Why is "Calculus graph element spotting" important?

Understanding the elements present in a graph of a calculus equation can help in solving and interpreting the equation, as well as making predictions about the behavior of the function. It is also a crucial skill in calculus-based courses and real-life applications.

3. What are some common elements to look for in a calculus graph?

Some common elements to look for in a calculus graph include the x- and y-intercepts, local and global extrema, inflection points, and asymptotes. These elements can provide valuable information about the behavior of the function and how it changes over its domain.

4. How can I improve my skills in "Calculus graph element spotting"?

Practice is key in improving skills in "Calculus graph element spotting". Familiarize yourself with different types of graphs and their corresponding equations, and work on identifying and analyzing their elements. You can also seek help from a tutor or attend review sessions to further enhance your skills.

5. Are there any resources available for "Calculus graph element spotting"?

Yes, there are many resources available online and in textbooks that provide practice problems and explanations for "Calculus graph element spotting". Some websites also offer interactive tools and simulations to help visualize and understand the concepts better. You can also consult with your teacher or professor for additional resources and guidance.

Similar threads

Replies
4
Views
2K
Replies
3
Views
2K
Replies
3
Views
737
Replies
3
Views
1K
Replies
24
Views
2K
Replies
11
Views
2K
Back
Top