Solving 2 Problems: Finding Critical Points & Differentiability

In summary, the conversation involves two problems. The first problem is to find and classify the critical points of a given function. The second problem is to show that a given function is differentiable everywhere, has a non-zero derivative, and maps no neighborhood around the zero point bijectively onto a neighborhood around the zero point. The discussion also includes an attempt at solving the first problem and a question about the second problem.
  • #1
gassi
3
0
I have two problems. I posted the first problem before but I still can´t solve it.

Homework Statement



Find and classify the critical points of f(x,y,z) = xy + xz + yz + x^3 + y^3 + z^3


Homework Equations



-

The Attempt at a Solution



df/dx = y + z +3x^2, df/dy = x + z + 3y^2 and df/dz = x + y + 3z^2

a point x is a critacal point if the gradient equals 0.

Obviously (0,0,0) is a critical point but I´m not sure how to find the others.

I know this is symmetric but I cant´t figure out were to go from here??

Here is the second problem:

I have a function from R to R, f(x) = x + 2*x^2*sin(1/x) if x is not 0 and f(x) = 0 if x is 0.
I´m supposed to show that this is differentable everywhere, that f'(x) is not 0 and that f maps no neighbourhood around the zero point bijective on a neighbourhood around the zero point.

I think I know how to show that this is differentable everywhere, that f'(x) is not 0 but I´m having difficulties with the last one.
 
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  • #2
In the first can I subtracct df/dx - df/dy and get y - x - 3x^2 - 3y^2 = 0 which gives
y(1-3y) = x(1-3x). Therefore x=y and then get that x=z=y??

I also have to tell wether they are maximum, minimum or saddle points.
 
Last edited:

1. What is the difference between critical points and differentiability?

Critical points are points on a function where the derivative is equal to zero or undefined. Differentiability refers to whether or not a function has a well-defined derivative at a given point. In other words, a function is differentiable at a point if the derivative exists at that point.

2. How do you find critical points and determine differentiability?

To find critical points, set the derivative of the function equal to zero and solve for the variable. To determine differentiability, check if the derivative exists at the critical points found. If the derivative exists at all critical points, the function is differentiable at those points.

3. Why are critical points and differentiability important in solving problems?

Critical points help us identify important features of a function, such as maximum or minimum values. Differentiability is important because it allows us to calculate the slope of the function at a given point, which is necessary for finding critical points and understanding the behavior of the function.

4. Can a function have critical points but not be differentiable?

Yes, a function can have critical points but not be differentiable. This can occur when the derivative is undefined or discontinuous at the critical point.

5. Are there any shortcuts or tricks for finding critical points and determining differentiability?

There are some techniques for finding critical points, such as the first and second derivative tests, which can help identify local maximum and minimum points. However, determining differentiability ultimately requires understanding the definition and properties of derivatives and critical points.

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