Find intervals of f increasing/decreasing

  • Thread starter phillyolly
  • Start date
  • Tags
    intervals
In summary: See my next post for how to find them.In summary, the function f(x) = x2/(x2+3) has a domain of all real numbers and is an even function. The intervals on which it is increasing are when x is greater than 0, and the intervals on which it is decreasing are when x is less than 0. The local minimum value of f(x) is 0 at x=0. To find the intervals of concavity, take the second derivative and determine where it is positive and negative. The inflection points are where the concavity changes, which in this case are at x=1 and x=-1.
  • #1
phillyolly
157
0

Homework Statement



(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of concavity and the inflection points.

x2/(x2+3)

Homework Equations


The Attempt at a Solution



a) I find that
f'(x)=6x/(x2+3)2

I am stuck because (x2+3)2 has no solutions and I don't know how to define x in order to proceed further.
If anybody can help me out understanding this, I will be very thankful.
 
Physics news on Phys.org
  • #2
Look at the function to see what it's it does. as [tex]x\rightarrow\pm\infty[/tex], [tex]f(x)\rightarrow 1[/tex]. Also note that f(x) is an even function. Your calculation shows that the one and only minimum is at x=0. This should allow you to answer the question.

Mat
 
  • #3
phillyolly said:

Homework Statement



(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of concavity and the inflection points.

x2/(x2+3)

Homework Equations





The Attempt at a Solution



a) I find that
f'(x)=6x/(x2+3)2

I am stuck because (x2+3)2 has no solutions and I don't know how to define x in order to proceed further.
Why are you just looking at the denominator? Since x2+3 has no real solutions, this means that the domain of f is the entire real line.

f is increasing on intervals for which f'(x) > 0, and is decreasing on intervals for which f'(x) < 0.

You're going to need the second derivative, too, in this problem.
phillyolly said:
If anybody can help me out understanding this, I will be very thankful.
 
  • #4
hunt_mat said:
Look at the function to see what it's it does. as [tex]x\rightarrow\pm\infty[/tex], [tex]f(x)\rightarrow 1[/tex]. Also note that f(x) is an even function. Your calculation shows that the one and only minimum is at x=0. This should allow you to answer the question.

Mat

1) How do you know it is an even function?
2) Why [tex]f(x)\rightarrow 1[/tex]?
 
  • #5
1) Because f(-x) = f(x) for all x
2) Because x2/(x2 + 3) = 1 - 3/(x2 + 3). As x --> inf, f(x) --> 1. As x --> -int, f(x) --> 1.
 
  • #6
So how can I find the intervals of concavity based on a second derivative?
 
  • #7
phillyolly said:
So how can I find the intervals of concavity based on a second derivative?

Take the second derivative and figure out where it's positive and where it's negative?
 
  • #8
I did so, and in this case, x=1. however, the graph is concave down at 0 and it has nothing to do with a point 1.
 
  • #9
phillyolly said:
I did so, and in this case, x=1. however, the graph is concave down at 0 and it has nothing to do with a point 1.

x=1 isn't an 'interval of concavity'. It's just a point. If you mean it's an inflection point, yes it is. But it's not the only one.
 

1. What does it mean for a function to be increasing or decreasing?

For a function to be increasing, it means that as the input values increase, the output values also increase. In other words, the graph of the function rises from left to right. Similarly, a decreasing function means that as the input values increase, the output values decrease and the graph of the function falls from left to right.

2. How do you find intervals of a function that is increasing or decreasing?

To find intervals of a function that is increasing or decreasing, we need to first take the derivative of the function. Then, we set the derivative equal to zero and solve for the input values. These input values will be the critical points of the function. We can then use the first or second derivative test to determine if the function is increasing or decreasing on each interval between the critical points.

3. What is the first derivative test?

The first derivative test is a method used to determine if a function is increasing or decreasing on a specific interval. It involves evaluating the derivative of the function at a specific point within the interval. If the derivative is positive, then the function is increasing on that interval. If the derivative is negative, then the function is decreasing on that interval.

4. Can a function be both increasing and decreasing?

No, a function cannot be both increasing and decreasing at the same time. However, a function can have both increasing and decreasing intervals. This means that on certain intervals the function is increasing, while on others it is decreasing.

5. How can finding intervals of a function that is increasing or decreasing be useful?

Knowing the intervals of a function that is increasing or decreasing can be useful in many ways. For example, it can help us identify maximum and minimum points of the function, determine the behavior of the function, and make predictions about the function's future values. It can also be useful in optimization problems, where we want to find the maximum or minimum value of a function within a specific interval.

Similar threads

  • Calculus and Beyond Homework Help
Replies
30
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
953
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
929
  • Calculus and Beyond Homework Help
Replies
9
Views
462
  • Calculus and Beyond Homework Help
Replies
10
Views
831
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top