How can I find the critical points of a function involving fractions?

In summary, the minima and maxima of a function can be found by setting the derivative equal to zero and solving for x. Another approach is to complete the square and use the vertex formula to find the minimum or maximum point. The points where the first derivative is zero are called critical points, and the points where the second derivative is zero are called inflection points.
  • #1
Дьявол
365
0
How will I find minima,maxima or both of one function (if they exist)? For example:

[tex]f(x)=x^2-3x+2[/tex];

[tex]f(x)=x^2-2x-x+2[/tex];

[tex]f(x)=x(x-2)-(x-2)[/tex];

[tex]f(x)=(x-2)(x-1)[/tex]

x1,2=1,2

Those are the zeroes of the function.

But how will I find the critical points (minima,maxima) of the function if [itex]x \in \mathbb{R}[/itex]
 
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  • #2
Дьявол said:
How will I find minima,maxima or both of one function (if they exist)?

The minima/maxima of a function is the point(s) where [tex]f'(x)=0[/tex]
 
  • #3
That's a quadratic function. Since it is concave upward it has a minimum but no maxiumum. The minimum is at the vertex you need to find the vertex. You could do that by completing the square but since you have already found the x-intercepts of the function, you can use the fact that, for a parabola with vertical axis, the vertex always lies half way between the x-intercepts.

Since this problem has nothing to do with linear algebra or abstract algebra, I am moving it to "General Mathematics".
 
  • #4
Take the derivative of the originial function, then find where the derivative equals zero. Then plug those points (they are x-values) into your original function, and that will be your Y-coordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.
 
  • #5
phyzmatix said:
The minima/maxima of a function is the point(s) where [tex]f'(x)=0[/tex]

Don't forget about inflection points!
 
  • #6
Riogho said:
Take the derivative of the originial function, then find where the derivative equals zero. Then plug those points (they are x-values) into your original function, and that will be your Y-coordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.
Do you mean like f(x)=(x-a)2+b ?

HallsofIvy, When I draw graphic of the function it is not parabola.

f(3)=2
f(2)=0
f(1)=0
f(0)=2
f(-1)=6
f(-2)=12
f(-3)=20
 
  • #7
f(x) has a relative minimum or maximum at either the endpoints of a region you're observing or when f'(x)=0. To determine whether it's a minimum, find when f''(x) is positive or negative; when f''(x) is positive, the graph is concave up, and when f''(x) is negative, the graph is concave down. Therefore, you have a relative extrema when f'(x)=0; it's a maximum if f''(x) is positive, or a minimum if f''(x) is negative.

F''(x)=0 are your points of inflection.
 
  • #8
Desh627 said:
f(x) has a relative minimum or maximum at either the endpoints of a region you're observing or when f'(x)=0. To determine whether it's a minimum, find when f''(x) is positive or negative; when f''(x) is positive, the graph is concave up, and when f''(x) is negative, the graph is concave down. Therefore, you have a relative extrema when f'(x)=0; it's a maximum if f''(x) is positive, or a minimum if f''(x) is negative.

F''(x)=0 are your points of inflection.


Eh? If inflection points are where f''(x) = 0 (which they are), what do you call points where f'(x) = 0, but is neither a maximum nor a minimum? For example, f(x) = x^3, which has no maximum or minimum has f'(0) = 0. What are those points called if it's not inflection points?
 
  • #9
Дьявол said:
How will I find minima,maxima or both of one function (if they exist)? For example:

[tex]f(x)=x^2-3x+2[/tex];

Дьявол said:
Do you mean like f(x)=(x-a)2+b ?

HallsofIvy, When I draw graphic of the function it is not parabola.

f(3)=2
f(2)=0
f(1)=0
f(0)=2
f(-1)=6
f(-2)=12
f(-3)=20
? It certainly is a parabola! Any quadratic function has a parabola as its graph. And the points (3,2), (2,0), (1,0), (-1,6), (-2,12), (-3,20) definitely lie on a parabola. The vertex is, as I said before, half way between 1 and 2, at 3/2. f(3/2)= 9/4- 9/2+ 2= -9/2+ 2= -5/2. f(x)= (x- 3/2)^2- 5/2.
 
  • #10
Tac-Tics said:
For example, f(x) = x^3, which has no maximum or minimum has f'(0) = 0. What are those points called if it's not inflection points?
Saddle points.

Edit
Perhaps that was too hasty. The problem is that the term inflection point describes a point where the second derivative reaches this. Inflection points obviously do not in general coincide with stationary points. A saddle point is by definition a stationary point that is not a local extremum.
 
Last edited:
  • #11
Tac-Tics said:
Eh? If inflection points are where f''(x) = 0 (which they are), what do you call points where f'(x) = 0, but is neither a maximum nor a minimum? For example, f(x) = x^3, which has no maximum or minimum has f'(0) = 0. What are those points called if it's not inflection points?

Inflections points are the points where the 2nd derivative equals zero, at those points the concavity will change.

Where the first derivative is zero are called critical points, those are the relative max's and mins of a function.
 
  • #12
Riogho said:
Where the first derivative is zero are called critical points
More appropriately, the points where the first derivative is zero are called stationary points. The absolute value function has a well-defined critical point: x=0. It has no stationary point because the derivative is non-zero everywhere except for x=0 and is not defined at x=0.
those are the relative max's and mins of a function.
Not all stationary points are extremal. Tac-Tics gave a perfectly good example of a function that has a stationary point but no extremal points.
 
  • #13
A quadratic function does not have any "inflection" points. And the original question has already been completely answered.
 
  • #14
Riogho said:
Where the first derivative is zero are called critical points

Ah, yes. Thank you. It's been about 6 years since I studied this. The concept is there, but the terminology is fuzzy and needs dusting :-)
 
  • #15
Since all of your functions are quadratic, an alternative to calculus is to complete the square. Ie., your first function is f(x) = x^2 - 3x + 2 = (x - 3/2)^2 - 5/2. Since squares of real numbers are never negative, your function has the minimum point (3/2, -5/2).
 
  • #16
Thanks for the posts. I found the minima with derivation of the original formula. So my goal is [itex]x^2-3x+2=(x-a)^2+b[/itex]
[tex]x^2-3x+2=x^2-2ax+a^2+b[/tex]
[tex]-3x+2=-2ax+(a^2+b)[/tex]
[tex]-3=-2a[/tex]
and
[tex]2=(a^2+b)[/tex]

So, a= 3/2
and
[tex]b=2-9/4[/tex]
[tex]b=-1/4[/tex]

[tex] x^2-3x+2=(x-3/2)^2-1/4 \geq -1/4=f(3/2)[/tex]

So 3/2 is the local minimum.
 
  • #17
In other words, the vertex of the parabola is at x= 3/2, half way between the x-intercepts. That's what I said in post 3.
 
  • #18
I am again stuck with minima and maxima of function.

The original problem:
[tex]f(x)=\frac{4x}{3x^2+4}[/tex]

Now, I tried:

[tex]f(x)=\frac{4}{3}*\frac{x}{x^2+\frac{4}{3}}=\frac{4}{3}*\frac{1}{x+\frac{4}{3x}}[/tex] and I don't know how to go on out of here. I was trying to have something in the form (x+a)2+b > or < b

Also I tried [tex]\frac{4x}{3x^2+4} + 1 - 1=\frac{3(x^2+\frac{4}{3}x+\frac{4}{3})}{3(x^2+\frac{4}{3})} - 1 = \frac{(x-\frac{2}{3})^2+\frac{6}{9} \geq \frac{6}{9}}{(x-0)^2 + \frac{4}{3} \geq \frac{4}{3} } - 1[/tex]
Again nothing. Please help me!

Thanks in advance.
 

1. What is a minima of a function?

A minima of a function is the lowest point or value of the function, also known as the global minimum. It is the point where the function starts to increase again after decreasing.

2. How is a maxima of a function defined?

A maxima of a function is the highest point or value of the function, also known as the global maximum. It is the point where the function starts to decrease again after increasing.

3. How do you find the minima and maxima of a function?

The minima and maxima of a function can be found by taking the derivative of the function and setting it equal to zero. Solving for the values of x that make the derivative zero will give the x-coordinates of the minima and maxima points. These points can then be plugged back into the original function to find the corresponding y-coordinates.

4. Can a function have multiple minima and maxima?

Yes, a function can have multiple minima and maxima. These points are called local minima and maxima, and they occur when the function changes direction from increasing to decreasing or vice versa. Local minima and maxima can also be found by taking the second derivative of the function and determining the sign of the derivative at those points.

5. How are minima and maxima used in real-world applications?

Minima and maxima of functions are used in various fields such as economics, physics, and engineering. They are used to optimize functions and find the most efficient solution to a problem. For example, in economics, minima and maxima are used to find the minimum cost or maximum profit for a company. In physics, they are used to find the minimum energy state of a system.

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